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PLANE   GEOMETRY 


A  SERIES  OF  MATHEMATICAL  TEXTS 

EDITED    BY 

EARLE  RAYMOND  HEDRICK 


THE  CALCULUS 

By   Ellery    Williams    Davis    and    William    Charles 
Brenke. 

PLANE   AND    SOLID   ANALYTIC   GEOMETRY 

By  Alexander  Ziwet  and  Louis  Allen  Hopkins. 

PLANE     AND     SPHERICAL     TRIGONOMETRY     WITH 
COMPLETE   TABLES 
By  Alfred  Monroe  Kenyon  and  Louis  Ingold. 

PLANE     AND     SPHERICAL     TRIGONOMETRY     WITH 
BRIEF   TABLES 
By  Alfred  Monroe  Kenyon  and  Louis  Ingold. 

THE   MACMILLAN   TABLES 

Prepared  under  the  direction  of  Earle  Raymond  Hedrick. 

PLANE   GEOMETRY 

By  Walter  Burton  Ford  and  Charles  Ammerman. 

PLANE   AND    SOLID   GEOMETRY 

By  Walter  Burton  Ford  and  Charles  Ammerman. 

SOLID   GEOMETRY 

By  Walter  Burton  Ford  and  Charles  Ammerman. 


PLANE  GEOMETRY 


BY 

WALTER  BURTON   FORD 

JUNIOR    PROFESSOR   OF   MATHEMATICS,    THE    UNIVERSITY    OF    MICHIGAN 

AND 

CHARLES   AMMERMAN 

THE    WILLIAM    McKINLEY    HIGH    SCHOOL,    ST.    LOUIS 


EDITED    BY 

EARLE   RAYMOND   HEDRICK 


!C 


Neto  gork 

THE   MACMILLAN   COMPANY 

1920 

All  righU  reserved 


COPTEIGHT,    1913, 

By  the  MACMILLAN  COMPANY. 


-Set  dp  und'electrotyped.    Published  September,  1913. 


J.  S.  Gushing  Co.  —  Berwick  &  Smith  Co. 
Norwood,  Mass.,  U.S.A.        • 


PREFACE 

While  the  preparation  of  a  textbook  on  Geometry  presents 
much  the  same  problem  whenever  undertaken,  the  arrangement 
of  its  details  and  the  balancing  of  its  parts  must  be  expected 
to  change  as  new  educational  ideals  become  established,  and  as 
new  conditions  of  society  arise. 

It  is  no  longer  possible,  for  example,  to  expect  favor  for  the 
typical  lengthy  text  of  a  generation  ago.  Although  such  texts 
proceed  in  good  logical  fashion  to  deduce  the  traditional  theo- 
rems one  after  another,  yet  they  make  little  or  no  appeal  to  the 
world  of  common  experience ;  and  they  present  far  more  mate- 
rial than  the  student  can  absorb  in  the  course  as  it  is  now 
usually  taught.  Neither  is  the  ideal  book  something  so  far 
removed  from  this  traditional  type  that  its  chief  feature  is  a 
large  variety  of  illuminating  diagrams  dra\vn  from  the  Arts. 

Between  these  two  extremes  lies  the  only  proper  course 
to-day,  and  this  the  authors  have  tried  diligently  to  find.  The 
traditional  manner  of  presentation  in  a  logical  system  is  pre- 
served, but  logical  development  is  not  made  the  sole  purpose 
of  the  book.  Thus,  problems  drawn  from  the  affairs  of  prac- 
tical life  are  inserted  in  considerable  number.  The  function  of 
such  problems  is  not  to  train  the  student  in  the  technique  of 
any  of  the  Arts ;  rather  it  is  to  illuminate  the  geometric  facts, 
and  to  make  clear  their  importance  and  their  significance. 

Geometry  is  and  is  likely  to  remain  primarily  a  cultural, 
rather  than  an  informational  subject.  But  the  intimate  con- 
nection of  Geometry  with  human  activities  is  evident  ui)on 
every  hand,  and  constitutes  fully  as  much  an  integral  part  of 
the  subject  as  does  its  older   logical   and   scholastic  aspect. 

459955 


vi  PREFACE 

Teachers  of  Geometry  throughout  the  country  recognize  this 
and  appreciate  strongly  the  value  of  the  study,  not  only  as 
subject-matter  in  which  practice  in  logical  processes  is  ideally 
simple  and  beautiful,  but  also  as  an  instrument  for  meeting 
real  human  needs.  It  is  certainly  most  desirable  from  the 
standpoint  of  effective  instruction  that  the  students  also  come 
to  realize  all  of  these  merits,  thus  answering  for  themselves 
the  persistent  question  as  to  why  the  subject  is  studied.  The 
necessary  attention  and  concentration  of  mind  on  the  students' 
part  will  then  become  from  the  very  outset  easy  and  natural. 

Attention  is  called  to  the  Introductory  chapter.  In  it  the 
student  is  acquainted  with  the  use  of  the  ruler  and  compasses 
in  the  more  simple  construction  problems,  and  is  thus  brought 
early  face  to  face  with  many  of  the  fundamental  notions  of 
Geometry.  Great  care  has  been  exercised  at  this  point  that 
the  student  may  approach  the  more  rigid  demands  of  the  de- 
ductive side  of  the  subject  gradually,  rather  than  plunge  into 
them  at  the  outset.  Even  in  Chapter  I,  in  which  strictly 
deductive  work  commences,  no  abrupt  change  of  style  occurs, 
but  an  easy  transition  is  made  into  the  parts  that  so  often 
appear  foreign  and  unintelligible  to  the  beginner.  Indeed, 
throughout  the  book,  an  effort  has  been  made  to  soften  the 
austere  and  unnatural  style  that  has  frequently  proved  a 
bar  to  ready  comprehension,  and  to  avoid  an  excess  of  sym- 
bolic characters  and  technical  phrases  that  do  not  add  to  the 
reasoning. 

The  book  is  distinguished  by  its  acceptance  of  the  principle 
of  emphasis  of  important  theorems  laid  down  by  the  Committee 
of  Fifteen  of  the  National  Education  Association  in  their  Re- 
port.* Thus,  theorems  of  the  greatest  value  and  importance 
are  printed  in  bold-faced  type,  and  those  whose  importance  is 
considerable  are  printed  in  large  italics. 


*  Printed  as  a  separate  pamphlet  with  the  Proceedings  for  1912.    Reprinted 
also  in  School  Science,  1911,  and  in  The  Mathematics  Teacher,  December,  1912. 


PREFACE  vii 

The  Report  just  mentioned  has  been  of  great  assistance,  and 
its  principles  have  been  accepted  in  general,  not  in  a  slavish 
sense  but  in  the  broad  manner  recommended  by  the  Committee 
itself.  A  perusal  of  the  Report  will  give,  more  fully  and 
accurately  than  could  be  done  in  this  brief  preface,  the  consid- 
erations which  led  to  the  adoption  of  these  principles,  in  par- 
ticular, the  principle  of  emphasis  upon  important  theorems, 
both  by  the  Committee  and  by  the  authors  of  this  book. 

The  authors  are  indebted  also  to  many  another  book  and  to 
many  recent  reports  and  papers  for  ideas  and  problems.  In 
the  Introduction,  the  effect  of  the  excellent  little  book  of 
Godfrey  and  Siddons  *  is  sufficient  to  deserve  explicit  mention. 

Detailed  description  of  the  principles  followed  in  the  Solid 
Geometry  may  be  omitted,  since  they  are  in  spirit  the  same  as 
those  of  the  Plane  Geometry.  The  great  excellence  of  the 
figures,  particularly  that  of  the  very  unusual  and  effective 
*  phantom'  halftone  engravings  in  the  Solid  Geometry,  de- 
serves mention.  These  figures  should  go  far  toward  reliev- 
ing the  unreality  which  often  attaches  to  the  constructions  of 
Solid  Geometry  in  the  minds  of  students. 

W.  B.  FORD. 
CHARLES  AMMERMAN. 
E.  R.  HEDRICK,  Editor. 


♦  Godfrey  and  Siddons,  Plane  Geometry,  Cambridge  University  Press. 


CONTENTS 


Introduction 

Part  I.       Drawing  Simple  Figures 
Part  II.     The  Principal  Ideas  used  in  Geometry- 
Part  III.   Statements  for  Reference 
Supplementary  Exercises 

Chapter  I.    Rectilinear  Figures 

Part  I.       Triangles 

Parallel  Lines . 
Angles  and  Triangles 
Quadrilaterals 
Polygons 
Locus  of  a  Point 


Part  II. 

Part  III. 

Part  IV. 

Part  V. 

Part  VI. 

Miscellaneous  Exercises  . 


Chapter  II.    The  Circle 

Part  I.       Chords  —  Arcs  —  Central 
Part  II.     Tangents  and  Secants 
Part  III.   Measurement  of  Angles 
Part  IV.    Construction  Problems 
Miscellaneous  Exercises  . 


Angl 


Chapter  III. 
Part  I. 
Part  II. 
Part  III. 
Part  IV. 

Part  V. 


Proportion.    Similarity 

General  Theorems  on  Proportion 
Proportional  Line  Segments  . 
Similar  Triangles  and  Polygons 
Proportional  Properties  of  Chords,  Secants 

and  Tangents       .... 
Similar  Kight  Triangles,  Trigonomet 

tios 

Miscellaneous  Exercises 


Ra 


PAOBS 

1-33 

1-9 
10-25 
26-29 
30-33 

35-88 

35-48 
49-57 
58-67 
68-75 
76-78 
79-82 
83-88 

89-125 

89-98 

99-106 

107-116 

117-119 

120-125 

126-161 

126-129 
130-138 
139-147 

148-151 

152-156 
157-161 


CONTENTS 


IX 


PAGES 

Chapter  IV.    Areas  of  Polygons.    Pythagorean  Theorem  .     162-186 
Miscellaneous  Exercises .        .        .        .        .        .        .     180-185 

Chapter  V.    Regular  Polygons  and  Circles       .        .        .     186-205 
Miscellaneous  Exercises 201-205 

Appendix  to  Plane  Geometry:  Maxima  and  Minima        .     206-213 

Tables i-xxvii 

Table  I.       Quantities  Determined  by  a  Given  Angle    .  ii-vii 

Table  IF.     Powers  and  Roots viii-xxvi 

Table  III.   Important  Numbers xxvii 

Index xxix-xxxi 


?e 


This  edition  contains  only  the  Chap- 
ters relating  to  Plane  Geometry,  as 
listed  above. 

An  edition  that  contains  also  the 
usual  topics  of  Solid  Geometry  is 
published  under  the  title  Plane  and 
Solid  Geometry.  The  Solid  Geometry 
is  also  published  separately. 


PLAICE   GEOMETRY 

INTRODUCTION 
PART   I.     DRAWING   SIMPLE   FIGURES 

1.  Geometry  is  the  branch  of  mathematics  that  deals  with 
space.  In  it  we  consider  points,  lines,  triangles,  circles,  spheres, 
and  so  forth.  Such  questions  as  the  following  are  considered : 
When  do  two  triangles  of  different  shapes  have  the  same  area  ? 
How  does  the  length  around  a  circle  compare  with  the  diameter  ? 
How  can  the  volume  of  a  sphere  be  calculated  ?  Geometry 
not  only  answers  many  questions  of  this  kind,  but  it  also  pre- 
sents a  systematic  study  of  the  general  principles  involved, 
and  the  reasons  for  all  statements  are  given  with  care. 

2.  Ruler  and  Compasses.  In  studying  geometry  it  is  desir- 
able to  draw  accurate  figures.  The  student  will  need  a  nder^ 
with  which  to  draw  (or  extend)  straight  lines;  and  a  pair  of 
compasses,  with  which  to  draw  circles  and  to  transfer  lengths 
from  one  position  to  another. 


Fig.  1 

For  distinctness,  the  curve  that  forms  the  circle  is  often 
called  the  circumference  of  the  circle.  The  position  of  the 
fixed  point  of  the  compasses  is  called  the  center  of  the 
circle.  Any  portion  of  the  circumference  is  called  an  arc  of 
the  circle.  The  distance  from  the  center  to  the  circumference, 
which  is  the  same  all  the  way  around  the  circle,  is  called  the 
radius. 

B  1 


INTRODUCTION 


[§3 


3.  Problem  1.  Draw  with  a  ruler  and  compasses  a  triangle 
each  of  whose  sides  is  one  inch  long. 

[A  triangle  is  a  figure  bounded  by  three  straight  sides.] 

Solution.  With  a  ruler  and  pencil  draw  a  line  and  mark 
two  points  A  and  B  on  it  one  inch  apart. 

Around  each  of  these  points  as  a  center,  with  a  radius  one 
inch  long,  draw  an  arc  of  a  circle  above  the  line  AB  so  that 
the  two  arcs  cut  each  other  in  a  third  point,  which  we  will  call  C. 

C 


Fig.  2 

Connect  A  with  0  by  a  straight  line,  and  connect  B  and  C 
by  another  straight  line. 

The  triangle  ABC  is  the  required  figure ;  state  why  each 
side  is  one  inch  long. 

EXERCISES 

1.  Show  how  to  draw  a  triangle  each  of  whose  sides  is  a 
given  length  AB.       [Hint.     See  Problem  1.] 

2.  Copy  the  adjoining  figure. 
Draw  all  the  triangles  that  have 
equal  sides  which  you  can  by  draw- 
ing straight  lines  connecting  pairs 
of  points  marked  in  the  figure. 

3.  Draw  in  separate  positions 
on  heavy  paper  or  on  cardboard 
two  triangles,  each  of  whose  sides  is  three  inches  long.  Cut 
out  one  of  these.  Will  it  fit  precisely  upon  the  other  one  ? 
Will  it  fit  one  drawn  by  another  stude»i^ 


§4]  DRAWING  SIMPLE  FIGURES  3 

4.    Problem  2.     Draw  with  a  ruler  and  compasses  a  triangle 
ichose  three  sides  are  equal,  respectioely,  to  three  given  lengths. 

Solution.     Let  a,  6,  c  be  the  three  given  lengths.     (Fig.  3.) 


Draw  a  line.  Mark  two  points  A  and  B  on  it  so  that  the 
distance  AB  is  equal  to  one  of  the  given  lengths,  say  c.  This 
is  done  with  the  compasses. 

Around  A  as  center  draw  an  arc  of  a  circle  with  a  radius 
equal  to  another  of  the  given  lengths,  say  h. 

Around  B  as  center  draw  another  arc  with  a  radius  equal  to 
the  remaining  given  length  a,  and  in  such  a  position  that  the  two 
arcs  cut  each  other. 

Let  the  point  where  the  two  arcs  cut  each  other  be  called  C. 
Draw  CA  and  CB.     Then  the  triangle  ABC  is  the  one  required. 

EXERCISES 

1.  Draw,  with  ruler  and  compasses,  a  triangle  whose  three 
sides  are,  respectively,  two  inches,  three  inches,  and  four  inches 
in  length. 

2.  Is  there  any  difference  between  this  triangle  and  one 
whose  sides  are,  respectively,  four  inches,  two  inches,  and  three 
inches  in  length?  Draw  both  triangles  separately  on  card- 
board, cut  one  out,  and  see  if  it  will  exactly  fit  on  the  other 
one.     Was  it  necessary  to  turn  it  over  ? 

3.  What  can  you  say  about  any  two  triangles  if  the  three 
sides  of  one  are  equal,  respectively,  to  the  three  sides  of  the 
other  ? 


4  INTRODUCTION  [§5 

5.    Problem  3.     Drav^,  by  means  of  ruler  and  compasses,  a 
perpendicular*  to  a  given  straight  line  at  a  point  in  that  line. 


fP 


I  C 

Fig.  4 

Solution.  Let  AB  be  the  given  line  and  C  the  given  point 
in  it. 

With  C  as  center,  and  with  any  convenient  radius,  draw- 
arcs  of  the  same  circle  cutting  AB  at  two  points  which  we  will 
call  F  and  Q. 

With  a  somewhat  longer  radius  than  before,  draw  two  arcs 
of  circles  above  AB  with  centers  at  P  and  Q,  being  careful 
that  these  arcs  have  the  same  radius.  These  two  arcs  will  cut 
each  other  at  a  point  that  we  will  call  D. 

Draw  the  straight  line  joining  C  and  I).  This  is  the  required 
perpendicular  to  AB  at  O;  and  there  is  no  other. 

EXERCISES 

1.  Draw  with  ruler  and  compasses  a  perpendicular  to  a  line 
four  inches  long  at  a  point  one  inch  from  the  right-hand  end. 

2.  Draw  a  triangle  with  two  sides  perpendicular  to  each 
other,  making  the  perpendicular  sides  three  inches  long  and 
four  inches  long,  respectively.     Measure  the  third  side. 

[The  third  side  should  be  five  inches  long.  In  making  the  drawing, 
any  lines  may  be  extended  as  desired.] 


*  The  perpendicular  CD  can  be  drawn,  of  course,  with  a  carpenter's  square 
or  a  drawing  triangle ;  but  the  problem  is  to  draw  it  loith  ruler  and  compasses 
only. 

The  drawing  may  be  tested  by  folding  it  in  a  crease  along  the  line  CD  ; 
then  one  end  CB  of  AB  should  fall  on  the  other  end  CA  of  AB.  If  it  does, 
CD  is  said  to  be  perpendicular  to  AB  at  C. 


§6]  DRAWING  SIMPLE   FIGURES  5 

6.    Problem  4.     Draw  ivith  ruler  and  compasses  a  perpen- 
dicular to  a  given  straight  line  from  a  given  p>oint  not  on  that  line. 


^ 


L_^ 


Fig.  5 

Solution.  Let  AB  be  the  given  line  and  let  P  be  the  given 
point  not  on  that  line. 

With  P  as  center  and  with  any  radius  long  enough  to 
reach  past  the  line  AB,  draw  two  arcs  of  the  same  circle 
so  as  to  cut  the  line  AB  in  two  points,  which  we  shall  call 
R  and  S. 

Around  the  points  R  and  S  as  centers,  with  the  same  radius 
for  both,  draw  two  arcs  of  circles  below  AB,  and  let  C  be  the 
point  where  they  cut  each  other. 

The  straight  line  joining  P  and  C  is  the  desired  perpendicii- 
lar.  Show  this  by  folding  the  figure  on  the  line  PC  and  con- 
vincing yourself  that  TSB  falls  along  TRA. 

EXERCISES 

1.  Draw  a  triangle  whose  three  sides  are  in  the  ratios  2:3:3. 
Then  draw  with  ruler  and  compasses  a  perpendicular  from  each 
corner  to  the  opposite  side. 

[The  accuracy  of  the  drawing  can  be  tested  by  the  fact  that  these  three 
perpendiculars  should  meet  in  one  point.] 

2.  In  every  drawing  of  one  side  of  a  house,  a  line  is  drawn 
representing  the  horizontal  (or  base)  line.  Show  how  to  draw 
with  ruler  and  compasses,  through  any  point  in  the  figure,  a 
vertical  line,  that  is,  a  perpendicular  to  the  base  line. 


6  INTRODUCTION  [§  7 

7.    Problem  5.     Draw  with  ruler  and  compasses  an  angle  equal 
to  a  given  angle. 

[An  angle  is  formed  by  two  portions  of  straight  lines  that  end  at  the 
same  point.  ] 


B 

Fig.  6 

Solution.  Given  the  angle  at  A,  the  problem  is  to  draw  an 
angle  equal  to  A  from  a  point  P  on  another  line  MN. 

With  A  as  center,  draw  an  arc  of  a  circle  cutting  the  lines 
that  form  the  angle  J.  in  the  two  points  B  and  C. 

With  P  as  a  center,  draw  an  arc  x  with  the  same  radius  as 
that  just  used,  cutting  MN  at  Q  and  extending  upwards  quite 
a  distance. 

With  Q  as  center,  and  with  a  radius  equal  to  the  distance 
from  B  to  G,  draw  an  arc  y  cutting  the  arc  ic  at  a  point  0. 

Then  the  angle  at  P  formed  by  the  portions  of  straight  lines 
PQN  and  PO  is  equal  to  the  given  angle  at  A,  for  the  two 
figures  will  fit  each  other  exactly  if  one  is  placed  upon  the 
other  with  AdX  P  and  AB  along  PQN. 

EXERCISES 

1.  Draw  two  angles  that  are  exactly  equal  to  each  other. 
[Hint.    Draw  any  angle  and  then  draw  another  equal  to  it.] 

2.  Draw  a  triangle  with  one  angle  equal  to  the  angle  in 
Fig,  6  and  with  the  two  sides  that  form  that  angle  two  inches 
long  and  three  inches  long,  respectively.  Cut  out  your  tri- 
angle and  compare  it  with  those  made  by  other  students.  Will 
it  fit  exactly  on  theirs  ? 

3.  What  can  you  say  of  any  two  triangles  if  an  angle  and 
the  two  sides  that  include  it  in  one  triangle  are  equal  to  the 
corresponding  parts  of  the  other  triangle  ?     Why  ? 


§8]  DRAWING  SIMPLE  FIGURES  7 

8.    Problem  6.     Divide  a  portion  of  a  straight  line  into  two 
equal  parts. 


% 


Fig.  7 

Solution.  Let  A  and  B  be  two  points  on  a  straight  line. 
Draw  around  A  and  B  as  centers,  with  the  same  radius,  two 
circles.  These  circles  will  cut  each  other  in  just  two  points  P 
and  Q  if  the  radius  is  large  enough.  Only  small  arcs  of  the 
circles  are  shown  in  the  figure. 

The  straight  line  joining  Pand  Q  cuts  the  line  AB  at  a  point 
0.  This  point  0  divides  AB  into  two  equal  parts  AO  and  OB, 
for  the  whole  figure  may  be  folded  on  the  line  PQ  so  that  O 
remains  at  0,  and  B  falls  on  A. 

EXERCISES 

1.  Show  how  to  divide  a  line  into  four  equal  parts. 

2.  How  can  you  divide  accurately  the  ruled  page  of  a  note- 
book into  two  columns  of  equal  width  ? 

3.  Show  that  the  line  PQ  in  Fig.  7  is  also  a  perpendicu- 
lar to  the  line  AB  by  noting  the  effect  of  folding  the  figure 
along  PQ. 

4.  Draw  any  triangle;  then  find  by  ruler  and  compasses 
the  middle  point  of  each  of  the  sides.  Connect  each  corner 
of  the  triangle  to  the  middle  point  of  the  opposite  side  by  a 
straight  line. 

[Do  these  three  new  lines  meet  in  one  point  ?  The  accuracy  of  the 
drawing  may  be  tested  in  this  manner.] 


8  INTRODUCTION  [§  9 

9.    Problem   7.     Divide  a  given  angle  into  two  equal  parts. 


Solution.  Let  the  given  angle  be  the  angle  at  B  between  the 
portions  of  straight  lines  BC  and  BA. 

With  B  as  center  and  with  any  radius  draw  a  circle  that  cuts 
BC  at  L  and  BA  at  M. 

With  L  and  M  as  centers,  draw  two  arcs  of  circles  with  the 
same  radius  (of  any  convenient  size)  so  that  these  two  arcs  cut 
each  other  at  a  point  G. 

Then  the  straight  line  joining  B  and  G  divides  the  given 
angle  into  two  equal  parts.  Convince  yourself  of  this  by  think- 
ing of  folding  the  figure  on  the  line  BG. 

EXERCISES 

1.  Show  how  to  divide  a  given  angle  into  four  equal  parts. 

2.  Draw  any  triangle  and  divide  each  of  its  angles  into  two 
equal  parts.  State  anything  that  you  notice  about  the  way 
in  which  the  three  new  lines  meet  each  other.  Try  to  see 
whether  the  same  thing  happens  on  another  triangle  of  a  differ- 
ent shape.     State  all  your  conclusions  in  one  sentence. 

3.  Draw  two  perpendicular  lines ;  then  divide  the  angle  be- 
tween them  into  two  equal  parts.  State  the  connection  between 
this  problem  and  the  processes  that  occur,  for  example,  in 
making  the  corner  of  a  picture  frame,  or  in  mortising  a  joint. 
Can  you  mention  any  other  manufacturing  processes  in  which 
the  same  problem  arises  ? 


§9] 


DRAWING  SIMPLE  FIGURES 


9 


MISCELLANEOUS   EXERCISES   FOR  PART   I 

1.  A  shelf  is  to  be  fitted  into  a  corner 
near  a  window,  and  is  to  be  triangular. 
The  distance  from  the  corner  to  the  win- 
dow casing  is  9  inches,  the  shelf  is  to  be 
12  inches  long  on  the  other  wall  of  the 
corner,  and  the  edge  of  the  shelf  is  to  be 
15  inches  long.  Are  these  measurements 
enough  to  fix  the  shape  of  the  shelf  ? 
Why  ?  If  a  carpenter  cuts  out  of  a  piece 
of  board  a  triangle  that  has  exactly  these  lengths  of  sides,  why 
is  it  that  it  will  fit  in  the  place?  See  Problem  2,  p.  3,  and 
Ex.  3,  p.  3.     Will  the  shelf  fit  either  side  up? 

2.  Draw  a  triangle  with  each  side  just  one  third  the  length 
of  the  corresponding  side  of  the  triangle  mentioned  in  Ex.  1. 

3.  A  square  is  bounded  by  four  lines  of  equal 
length,  two  of  which  are  perpendicular  to  each 
other  at  each  corner.  Draw  a  square  whose 
sides  are  each  two  inches  long. 

[The  drawing  may  be  tested  by  the  fact  that  the 
two  straight  lines  joining  the  two  pairs  of  opposite  comers  should  be  equal 
in  length.] 

4.  If  the  Korth  and  South  line  is  shown  on  any  map  of  a  city, 
show  how  to  draw  the  East  and  West  line  through  any  point. 

5.  Show  how  to  draw  a  triangle  with  any  given  angle  and 
with  the  sides  that  form  that  angle  of  any  given  lengths.  Will 
two  triangles  fit  each  other  exactly  if  made  with  the  same  given 
angle  and  the  same  given  lengths  of  the  two  sides  ?  Must  one 
of  the  triangles  be  turned  over  before  they  will  fit  ? 

6.  If  two  triangles  are  drawn  with  two  angles  of  one  equal 
to  two  angles  of  the  other,  respectively,  will  one  necessarily  fit 
the  other  exactly  ?  What  more  is  needed  to  insure  that  the 
triangles  shall  fit  each  other  exactly  ? 


10  INTRODUCTION  [§  10 

PART  II.     THE  PRINCIPAL   IDEAS  USED  IN 
GEOMETRY 

10.  Solids,  Surfaces,  Curves,  Points.  A  limited  portion  of 
space,  such  as  that  bounded  by  a  sphere  or  a  cube,  is  called  a 
geometric  solid. 

In  geometry  we  have  nothing  to  do  with  the  material  of  the 
solid.  Thus  we  would  speak  of  the  space  occupied  by  this 
book  as  a  solid,  without  thinking  that  it  is  made  of  cloth 
and  paper.  Even  very  soft  objects,  such  as  a  cloud  or  a 
drop  of  water,  would  be  thought  of  in  geometry  only  with 
respect  to  the  form  of  the  portion  of  space  they  occupy,  and 
this  portion  of  space  would  be  called  a  geometric  solid,  even 
though  the  actual  object  is  not  solid  in  the  sense  of  being  hard 
to  bend  or  warp.  We  often  think  of  the  actual  object  as  taken 
away,  and  we  consider  the  portion  of  space  that  it  did  or  would 
occupy. 

For  every  solid  there  is  an  ideal  boundary  that  separates  the 
solid  from  the  rest  of  space;  this  ideal  boundary  is  called  a 
surfox^e. 

Surfaces  are  either  curved,  as  is  the  surface  of  a  sphere ;  or 
else  they  are  flat  in  every  direction,  that  is,  plane.  A  plane 
surface  is  simply  called  a  plane.  Thus,  each  of  the  faces  of  a 
cube  is  a  piece  of  a  plane. 

When  two  surfaces  cut  each  other,  their  common  points  form 
a  curve. 

Thus,  when  a  plane  cuts  a  sphere,  the  curve  formed  by 
their  common  points  is  a  circle.  Again,  when  two  planes 
cut  each  other,  the  curve  formed  is  the  simplest  possible 
curve  —  a  straight  line.  As  here,  we  shall  often  use  the 
word  curve,  in  general,  to  include  straight  lines  as  special 
curves. 

A  point  is  that  which  is  common  to  two  curves  which 
cut  each  other;  in  particular,  two  lines  cut  each  other  in  a 
point. 


§  12]  FUNDAMENTAL  IDEAS  11 

11.  Ideal  Nature  of  Geometric  Things.  It  should  be  noticed 
that  the  points  and  curves  and  surfaces  just  mentioned  are  ideal 
things  which  cannot  be  actually  made  out  of  materials.  Thus 
a  surface  must  be  thought  of  as  having  no  thickness  whatever. 
A  curve  or  a  straight  line  must  be  thought  of  as  having  length 
but  no  breadth  or  thickness.     A  point  has  only  a  position. 

While  we  cannot  manufacture  such  things  as  these,  every 
one  recognizes  that  they  can  be  thought  of  in  this  ideal  way. 
What  we  can  do  is  to  represent  them  very  nearly :  thus  a 
point  is  represented  by  a  dot,  a  curve  by  a  line  drawn  with 
pencil  or  chalk,  a  surface  by  paper  or  tin  or  some  other  very 
thin  substance. 

12.  Angles.  Two  portions  of  straight  lines  that  end  at  the 
same  point  form  an  angle.  The  two  lines  are  called  the  sides  of 
the  angle,  and  the  common  end-point  is  called  the  vertex  of  the 
angle.  The  length  of  the  sides  has  nothing  to  do  with  the  size 
of  the  angle,  which  depends  only  on  the  amount  of  the  opening. 

Strictly  speaking,  when  two  portions  of  straight  lines  thus 
end  at  a  common  point,  there  are  two  angles  formed :  thus  in  the 
figure,  the  portions  of  straight  lines  BA 
and  BC  may  be  thought  of  as  forming  the 
angle  marked  x,  or  they  may  be  thought 
of  as  forming  the  very  much  larger  angle 
marked  y  which  is  all  that  is  left  of  the  j,^^  y 

plane  if  the  angle  marked  x  is  taken  away. 
We  shall  always  understand  that  the  smaller  angle  x  is  in- 
tended in  such  a  case,  unless  the  contrary  is  stated. 

An  angle  may  be  read  in  any  one  of  three  ways: 

(1)  By  the  single  letter  at  the  vertex ;  as  the  angle  B. 

(2)  By  three  letters,  one  on  each  side  and  one  at  the  vertex ; 
as,  the  angle  ABC.  The  middle  letter,  here  J8,  is  always  the 
one  that  stands  at  the  vertex  of  the  angle. 

{3)  By  a  single  letter  placed  in  the  opening  of  the  angle,* 
as,  the  angle  x. 


12 


INTRODUCTION 


[§13 


13.  Measurement  of  Angles.  Units.  The  most  usual 
imit  of  angle  is  the  degree  (°).  It  is  formed  as  follows :  Divide 
the  circumference  of  any  circle  into  360  equal  parts  (or  arcs), 
and  then  join  the  ends  of  one  of  these  arcs  to  the  center  of  the 
circle  by  straight  lines ;  the  angle  thus  formed  at  the  center  is 
one  degree.  Thus  one  degree  is  one  three  hundred  sixtieth  of 
one  complete  revolution. 

The  degree  is  subdivided  into  60  equal  parts,  called  minutes, 
(').  The  minute  is  divided  into  60  equal  parts,  called  seconds  ("). 
Thus  an  angle  of  10  degrees,  20  minutes,  and  15  seconds  is  writ- 
ten 10°  20'  15". 

14.  The  Protractor.  A  protractor  is  an  instrument  for  meas- 
uring angles.  It  is  a  half -circle  made  of  cardboard,  celluloid,  or 
metal,  with  the  center  marked  at  0  (Fig.  10)  and  with  the  circum- 


Protractor. 


ference  divided  by  fine  lines  into  180  equal  arcs.  Each  of  these 
arcs  corresponds  to  1°,  if  the  vertex  of  the  angle  is  placed  at  0. 
To  measure  an  angle,  place  the  protractor  upon  it  so  that  one 
side  of  the  angle  lies  along  the  radius  OA,  with  the  vertex  of 
the  angle  at  0.  Then  the  other  side  of  the  angle  will  fall  in 
some  such  position  as  OP,  and  the  number  of  degrees  and  frac- 
tions of  a  degree  can  be  read  off  directly  from  the  scale 


§14]  FUNDAMENTAL  IDEAS  13 

EXERCISES 

1.  Through  how  many  degrees  does  the  minute  hand  of  a 
clock  turn  in  fifteen  minutes  ? 

2.  Through  how  many  degrees  does  the  hour  hand  of  a  clock 
turn  in  one  hour  ? 

3.  The  second  hand  of  a  watch  turns  on  a  circular  dial  that 
is  divided  into  sixty  equal  parts.  What  is  the  angle  between 
two  successive  marks  ?  What  is  the  angle  between  the  mark  for 
10  seconds  and  the  mark  for  15  seconds  ?  What  is  the  angle 
between  the  mark  for  10  seconds  and  that  for  20  seconds  ? 

4.  Ordinary  scales  for  weighing  small  objects  are  often 
made  with  a  circular  face  like  a  clock  face ;  the  divisions  of  the 
scale  indicate  pounds ;  if  the  entire  face  represents  24  pounds, 
what  is  the  angle  between  two  successive  pound  marks  ? 

5.  There  being  16  ounces  in  one  pound,  what  is  the  angle 
between  two  successive  ounce  marks  on  the  scale  of  Ex.  4  ? 

6.  How  long  does  it  take  the  minute  hand  of  a  clock  to  turn 
through  36°  ?  How  long  does  it  take  the  hour  hand  of  a  clock 
to  turn  through  36°  ?     60°  ?     75°  ? 

7.  What  weight  will  cause  the  hand  of  the  scale  described 
in  Ex.  4  to  turn  through  15°  ?     60°  ?     75°  ?     150°  ? 

8.  Through  how  many  degrees  does  a  screwdriver  turn  in 
half  a  revolution  ? 

9.  If  a  wheel  makes  ten  revolutions  per  minute,  through 
how  many  degrees  does  it  turn  in  one  second  ? 

10.  Draw  an  angle  of  150°  with  a  protractor,  and  divide  it 
into  four  equal  parts  by  means  of  ruler  and  compasses.  Meas- 
ure the  resulting  angles  with  the  protractor:  How  much  error 
did  you  make  in  each  case  ? 

11.  Draw  an  angle  as  nearly  equal  to  75°  as  you  can  judge 
by  your  eye.  Then  measure  your  angle  with  a  protractor.  How 
much  error  did  you  make  ?    What  fraction  of  75°  is  your  error  ? 


14 


INTRODUCTION 


[§15 


15.  Generation  of  Curves  and  Surfaces  by  Motion.     We 

may  think  of  curves  and  surfaces  as  formed,  or  generated,  by 
motion. 

If  a  point  moves,  its  path  is  a  curve. 

If  a  curve  moves,  it  generates  a  surface. 

If  a  surface  moves,  it  generates  a  solid. 

Thus,  the  point  of  the  compasses  that  draws  a  circle 
may  be  thought  of  as  a  moving  point  that  is  generating 
the  circle.  Again,  if  a  circle  is  rotated  about  a  line  through 
its  center,  it  generates  the  surface  of  a  sphere.  If  a  square 
that  lies  horizontally  is  lifted  vertically,  it  generates  a  rec- 
tangular block ;  the  block  becomes  a  cube  when  the  height 
through  which  the  square  is  lifted  becomes  equal  to  one  of 
its  sides. 

Notice,  however,  that  a  moving  curve  does  not  always  gen- 
erate a  surface.  Thus,  when  a  wheel  turns  on  its  axle,  the 
curve  formed  by  its  circumference  is  moving,  but  no  surface  is 
being  generated.  Likewise,  a  surface  that  merely  slides  upon 
itself  does  not  generate  a  solid.  All  such  motions  as  these  are 
exceptions  to  the  general  rules  stated  above. 

16.  Generation  of  Angles  by  Rotation.  An  angle  may 
always  be  thought  of  as  generated  by  a  line  which  rotates  about 
one  of  its  extremities  (re- 
garded as  fixed).  Thus,  if  the 
line  AB  rotates  about  the 
point  A,  it  takes  one  after 
another  the  positions  AC,  AD, 
AE,  AF,  AG,  and  finally 
comes  back  to  its  original 
position  AB.  In  each  case 
it   makes   an  angle   with   its  Pj^  j^ 

original  position  AB,  and  it 

is  to  be  noted   that   this    angle    increases   as    the    rotation 

goes  on. 


t^ 


X 


§17] 


FUNDAMENTAL   IDEAS 


15 


|D   Right 
Angle 


17.  Important  Special  Angles.  An- 
other important  unit  angle  is  the  right 
angle,  which  is  the  angle  between  two 
lines  that  are  perpendicular  to  each 
other  (see  footnote,  p.  4). 

Still  another  important  angle  is  a 
complete  revolution,  the  angle  formed  when  a  line  turns  around 
one  of  its  extremities  until  it  comes  to  its  original  position. 
A  complete  revolution  is  also  called 
simply  a  revolution ;  or  sometimes  a 
•perigon.  ^       -^^^  A 

When  the  two  sides  of  an  angle  lie 
along  the  same  straight  line,  and  in  opposite  directions  from 
the  vertex,  the  angle  is  called  a  straight  angle. 

A  straight  angle  is  equal  to  two  right  angles,  since  a  perpen- 
dicular to  a  straight  line  makes  two  equal  right  angles  on  one 

side  of  the  line  to  which  it  is  per-    ^ 

^.      -  Straight 

pendicular.  Angif       ^— ^180 


B 
Fig.  12 


One  Revolution 


B 
Fig.  14 


A    revolution    is    equal    to    two      C — 
straight   angles,    or   to   four    right 
angles. 

Since  a  revolution  is  360°,  a  straight  angle  is  180°,  and  a 
right  angle  is  90°. 

An  acute  angle  is  an  angle  less  than  a  right  angle.  An 
obtuse  angle  is  an  angle  greater  than  a  right  angle  and  less 
than  a  straight  angle. 


Acute  Angle  Obtuse  Anglr 

Fig.  15 

If  two   straight  lines    cross   each    other   the    sum   of  the 

two  angles  formed  on  the  same  side  of  either  of  the  lines 

is   a  straight  angle.     If  one  of  them  is  acute,  the  other   is 

obtuse. 


16  INTRODUCTION  [§  17 

EXERCISES 

1.  In  Fig.  11,  pick  out  a  right  angle ;  an  acute  angle ;  an 
obtuse  angle;  an  angle  of  one  revolution;  a  straight  angle. 
Are  there  any  other  right  angles  in  the  figure  ? 

2.  Show  from  the  footnote  on  p.  4,  regarding  a  perpen- 
dicular, that  if  two  lines  cross  at  right  angles,  any  two  of  the 
four  angles  formed  are  equal  to  each  other. 

3.  How  many  degrees  are  there  in  half  a  right  angle  ?  in 
one  third  of  a  right  angle  ?  in  one  fourth  of  a  right  angle  ? 

4.  How  many  degrees  are  there  in  one  and  one  half  right 
angles  ?  in  two  right  angles  ?  What  is  another  name  for  two 
right  angles  ? 

5.  How  many  degrees  are  there  in  one  revolution  ?  in  one 
fourth  of  a  revolution  ?  in  one  twelfth  of  a  revolution  ?  in 
one  fifteenth  of  a  revolution  ? 

6.  Is  one  fifth  of  a  revolution  an  acute  or  an  obtuse  angle  ? 

7.  Is  two  thirds  of  a  straight  angle  acute  or  obtuse  ? 

8.  Name  two  streets  that  you  know  which  meet  each  other 
at  right  angles.  Name  two  that  do  not.  In  the  latter  case 
describe  the  corner  or  corners  at  which  there  is  an  acute  angle ; 
those  at  which  there  is  an  obtuse  angle. 

9.  Does  a  rafter  of  the  roof  of  a  barn  make  an  acute  or  an 
obtuse  angle  with  an  upright  in  the  side  wall  ?  What  can  you 
say  of  the  angle  at  the  peak  of  the  roof  where  the  rafters  join? 

10.  Through  what  kind  of  an  angle  has  a  door  turned  on  its 
hinges  when  it  is  said  to  be  ajar?  Can  a  door  be  opened 
through  an  obtuse  angle  ? 

11.  The  earth  turns  on  its  axis  once  in  24  hours.  How 
many  degrees  of  longitude  correspond  to  1  hour  ? 

12.  Apply  the  construction  for  dividing  an  angle  into  two 
equal  parts  (§  9)  to  a  straight  angle.  Show  that  the  construc- 
tion that  results  is  the  same  as  that  of  §  6. 


§19] 


FUNDAMENTAL  IDEAS 


17 


Fiu.  16 

Thus  the  angles  x  and  y 
C 


B 
Fig.  17 


18.  Relations  between  Two  Angles.  Two  angles  that  have 
a  common  vertex  and  one  common  side  between  them  are  called 
adjacent  angles.  Thus  the  angles  a  and  b  in 
the  hgure  are  adjacent  angles. 

If  the  sura  of  two  angles  is  equal  to  a  right 
angle,  the  two  angles  are  said  to  be  comple- 
mentary to  each  other;  or,  either  of  them  is 
called  the  complement  of  the  other. 
in   Fig.  17    are   complementary   to 
each  other. 

If  the  sura  of  two  angles  is  equal 
to  two  right  angles,  the  two  angles 
are  said  to  be  supplementary  to  each 
other ;  or,  either  of  them  is  called 
the  supplement  of  the  other.  Thus 
the  angles  x  and  y  in  Fig.  18  are 
supplementary  to  each  other. 

By  the  sum  of  two  angles  is  meant  the  angle  formed  by 
placing  the  angles  adjacent  to  each 
other;  the  sura  is  the  total  angle 
thus  formed,  as  in  the  preceding 
figures.  The  measure  of  the  sum 
in  degrees  is  the  sura  of  the  nuraber 
of  degrees  in  the  two  given  angles.  Fia.  18 

19.  Vertical  Angles.     If  two  lines  AB  and  CD  cross  each 
other  at  a  point  0,  the  angles  that  lie 
opposite  each  other  across  the  common 
vertex  are  called  vertical  angles. 

Thus,  in  Fig.  19,  the  angles  x  and  y 
are  vertical  angles ;  and  u  and  v  also 
are  vertical  angles.  Since  the  sura 
of  u  and  a;  is  a  straight  angle,  and  the 
sum  of  u  and  ?/  is  a  straight  angle,  it  is  easy  to  see  that  x  and  y 
must  be  equal.  That  is,  any  two  vertical  angles  are  equal  to 
each  other. 


Fig.  19 


18  INTRODUCTION  [§  19 

EXERCISES 

[In  this  list,  and  hereafter,  the  sign  Z  is  used  for  the  word  angle.l 

1.  Angle  ^5(7  is  a  right  angle.   IfZa;=40°,     C 
how  many  degrees  in  Z.y?     What  is  the  com- 
plement of  40°  ? 

2.  What  is  the  complement  of  30°  ?  50°  20'  ? 
45°  18' 20"?   74°  31'  14"? 

3.  The  complement  of  a  certain  angle  x  is 
2  X.     How  many  degrees  are  there  in  a;  ? 

4.  The  complement  of  a  certain  angle  is  eight 
times  itself.     What  is  the  angle  ?     Draw  a  diagram  by  means 
of  a  protractor. 

5.  Inthefiguve,  Z  ABC -hZCBD 
=  2  right  angles,  or  the  straight 
angle  ABD.    If  Z  ic=50°,  how  many 

degrees  are  there  in  Z  y?     What  is   D 

the  supplement  of  50°  ?  B  " 

6.  What  is  the  supplement  of  80°  ?  40°  15'  ?  100°  30'  20''  ? 

7.  The  supplement  of  a  certain  angle  a;  is  4  a;.     How  many 
degrees  are  there  in  a;  ?     Draw  a  diagram. 

8.  The  supplement  of  a  certain  angle  is  eleven  times  itself. 
What  is  the  angle  ?     Draw  a  diagram. 

9.  Compare  the  complements  of  two  equal  angles. 

10.  Compare  the  supplements  of  two  equal  angles. 

11.  What  kind  of  an  angle  is  equal   to  its  supplement  ? 

12.  Find  two  complementary  angles  whose  difference  is  36°. 

13.  Two  supplementary  angles  are  such  that  one  is  40°  more 
than  the  other.     Find  each  of  the  angles. 

14.  One  line  meets  another   line  so  that  one  angle  is  five 
times  its  adjacent  angle ;  find  each  of  the  angles. 

15.  How  do  two  angles  u  and  v  compare  if  they  have  che 
same  complement  ?  if  they  have  the  same  supplement  ? 


§21]  FUNDAMENTAL  IDEAS  19 

20.  Contrast  between  Drawing  and  Construction  of  Fig- 
ures. One  purpose  of  a  part  of  our  study  will  be  to  show 
how  figures  can  be  drawn  without  any  other  instruments  than  a 
ruler  and  compasses;  for  distinctness,  we  shall  say  that  a 
figure  has  been  constructed  when  only  these  instruments  have 
been  used.  The  direction  to  construct  a  figure  will  carry  with 
it  the  direction  that  only  these  instruments  are  to  be  used. 

There  is  no  objection  whatever  to  the  use  of  other  instru- 
ments for  quickly  sketching  a  figure.  Thus  perpendiculars 
may  be  drawn  by  means  of  a  fixed  square,  such  as  that  used 
by  carpenters  or  draftsmen.  We  shall  continue  to  use  the 
words  "to  draiv  ajigure^'  whenever  we  intend  that  other  draw- 
ing instruments  than  ruler  and  compasses  may  be  used. 

When  a  figure  is  to  be  drawn  only  in  a  very  rough  fashion, 
for  example  by  free-hand  without  any  instruments,  we  shall 
say  that  it  is  to  be  sketched. 

21.  Reduction  or  Enlargement  of  Drawings.  It  is  often 
inconvenient  or  impossible  to  draw  a  figure  its  actual  size,  or, 
as  is  often  said,  life-sized.  Thus  a  plan  of  a  house  cannot  con- 
veniently be  drawn  the  size  of  the  house. 

In  such  cases,  a  figure  is  drawn  in  which  every  distance  is 
reduced  in  the  same  ratio.  Thus  in  a  figure  drawn  half  size, 
the  distances  in  the  figure  are  all  half  the  actual  distances. 

House  plans  are  usually  drawn  on  a  scale  which  makes  a 
distance  of  one  quarter  of  an  inch  on  the  drawing  represent 
one  foot  in  the  actual  house ;  that  is,  the  scale  is  reduced  in 
the  ratio  of  one  to  forty-eight.  The  angles,  of  course,  remain 
unchanged. 

An  accurate  record  of  the  scale  used  should  be  written  on 
the  face  of  every  drawing  that  is  not  life-size.  Many  geomet- 
ric figures  do  not  change  their  properties  when  merely  en- 
larged, and  for  the  purpose  of  showing  something  about  a 
figure,  it  may  be  drawn  in  any  convenient  size.  Thus  angles 
are  not  changed  by  reducing  the  size  of  the  drawing. 


20  INTRODUCTION  [§  21 

Figures  described  in  exercises  in  which  small  distances  are 
used  should  be  enlarged  when  they  are  drawn  on  the  black- 
board. When  large  distances  are  mentioned  in  exercises,  the 
size  may  be  reduced  for  a  drawing  on  paper  or  at  the  board. 

EXERCISES 

1.  What  distance  on  the  drawing  represents  20  ft.  in  a 
house  plan  drawn  to  the  scale  mentioned  in  §  21  (^  in.  to  1  ft.)  ? 

2.  What  actual  distance  does  6  in.  represent  in  a  house  plan 
drawn  to  the  scale  mentioned  in  §  21  ? 

3.  What  are  the  real  dimensions  of  a  room  that  appears  on  a 
house  plan  to  be  21  by  3  in.,  if  the  plan  is  drawn  on  the  scale 
described  in  §  21  ?  What  is  the  actual  floor  area  of  the  room  ? 
(The  area  is  the  length  times  the  breadth.) 

4.  A  table  2  ft.  6  in.  wide  and  4  ft.  long  is  to  be  placed  in 
one  of  the  rooms.  How  large  a  spot  will  it  represent  in  the 
drawing  on  the  scale  of  §  21  ? 

5.  On  a  map  whose  scale  is  37  mi.  to  the  inch,  the  dis- 
tance between  Chicago  and  Ann  Arbor  is  5f  in.  What  is  the 
actual  distance  ? 

6.  New  York  is  143  mi.  from  Albany.  How  far  apart  are 
they  on  the  map  referred  to  in  Ex.  5  ? 

7.  A  ship  on  leaving  port  sails  N.W.  18  mi.,  then  N.  15 
miles.  Draw  a  map  showing  her  course,  using  a  scale  of  1 
in.  for  10  mi.  In  this  manner  find  (by  measurement  on 
your  map)  her  approximate  distance  and  her  bearing  from 
port ;  that  is,  how  many  degrees  West  of  North. 

8.  When  a  vertical  pole  20  ft.  high  casts  a  shadow  35  ft. 
long,  what  is  the  acute  angle  the  sun's  rays  make  with  the  hor- 
izontal ? 

[This  angle  is  called  the  angle  of  elevation  of  the  sun.  Draw  a  map, 
scale  10  ft.  to  the  inch,  and  use  a  protractor  to  measure  the  angle.] 


§231 


FUNDAMENTAL  IDEAS 


21 


Fig.  20 


Fig.  21 


22.  Triangles.  Notation.  A  figure  bounded  by  three 
straight  lines  is  called  a  triangle.  The  bounding  lines  are 
called  the  sides,  and  the  points  where  the  sides  meet  are  called 
the  vertices.  Usually,  the  small  letters,  a,  6,  c,  are  used  to  denote 
the  sides,  while  the  capital  letters.  A,  B,  C,  are  used  to  denote 
the  vertices.  The  side  a  is  then  always 
placed  opposite  the  vertex  A,  while  b  is 
likewise  placed  opposite  to  B,  and  c 
opposite  to  C,  as  indicated  in  the  figure. 

The  angle  at  A  is  called  the  included 
angle  of  the  sides  b,  c.     Similarly,  B  is  the  included  angle  of 
the  sides  a,  c ;  and  C  is  the  included  angle  of  a,  b. 

The  angles  at  A,  B,  C  are  known 
as  the  interior  angles  of  the  tri- 
angle. Besides  these,  every  tri- 
angle has  what  are  known  as 
exterior  angles.  In  the  figure,  x 
represents  the  exterior  angle  at  B.  In  general,  an  exterior  angle 
is  one  which,  like  x,  is  formed  between  one  side  of  the  triangle 
and  the  prolongation  (extension)  of  another  side. 

23.  Circles.  A  circle  is  a  closed  curve,  every  point  of  which 
is  equally  distant  from  a  fixed  point  within 

called  the  center.  The  distance  from  the  center 
to  any  point  on  the  circle  is  called  the  radius. 
A  straight  line  through  the  center  terminated 
by  the  circle,  is  called  a  diameter. 

It  follows  from  this  definition  of  a  circle 
that  all  its  radii  are  equal.  How  do  the 
radius  and  the  diameter  compare  in  length  ? 

When  several  circles  have  the  same  center,  but  different 
radii,  they  are  called  concentric.     Draw  three  concentric  circles. 

The  curve  that  forms  the  circle  is  often  called  the  circumference. 
The  word  circle  is  sometimes  used  to  denote  the  space  enclosed 
by  the  circumference. 


Fig.  22 


22  INTRODUCTION  [§  24 

24.  Squares  and  Rectangles.  Any  figure  bounded  by  four 
lines  is  called  a  quadrilateral. 

A  rectangle  is  a  quadrilateral  each  of  whose  angles  is  a  right 
angle.      The   two   sides   that  meet   at  any 
corner  (vertex)  of  a  rectangle  are  therefore 
perpendicular  to  each  other. 

A  line  that  joins  opposite  corners  (ver- 
tices) of  a  rectangle  is  called  a  diagonal. 
Thus  ^O  in  Fig.  23   is  a  diagonal  of  the  Fig.  23 

rectangle  A  BCD. 

If  all  the  sides  of  a  rectangle  are  of  equal  length,  it  is  called 
a  square. 

25.  Areas.  To  measure  an  area  of  any  sort,  a  unit  is  usually 
chosen  which  is  a  square,  any  side  of  which  is  equal  to  the 
unit  of  length.  The  most  common  unit  of  area  is  one  square 
foot ;  that  is,  a  square  each  of  whose  sides  is  1  ft.  long. 

Any  given  area  is  measured  by  comparing  its  size  with  that 
of  the  unit  square.  In  particular,  the  area  of  any  rectangle  is 
found  to  be  the  product  of  the  number  of  units  of  length  in 
its  base  times  the  number  of  units  of  length  in  its  height. 
This  rule  is  usually  learned  in  Arithmetic. 

Areas  that  are  bounded  by  curved  lines  or  by  pieces  of 
straight  lines  are  usually  measured  by  supposing  them  filled 
up  with  little  squares,  each  of  whose  areas  we  can  find.  If  the 
area  bounded  by  the  figure  cannot  be  precisely  filled  in  this 
way,  at  least  it  is  greater  than  the  sum  of  the  areas  of  those 
squares  that  lie  entirely  within  it ;  and  it  is  less  than  the  sum 
of  the  areas  of  squares  that  entirely  cover  it. 

A  good  practical  way  to  estimate  the  area  of  any  figure  is  to 
draw  it  on  paper  that  is  ruled  into  little  squares  of  known  size. 
Such  paper  (called  squared  paper,  or  cross  section  paper)  can 
usually  be  bought  at  any  stationers,  the  ruling  being  into 
squares  one  tenth  of  an  inch  on  each  side.  There  are,  of 
course,  one  hundred  such  squares  in  one  square  inch. 


§25] 


FUNDAMENTAL  IDEAS 


23 


Beneath,  in  Fig.  24,  several  figures  are  drawn  on  a  sheet  of  squared 
paper.  Estimate  the  areas  of  each  of  them  in  the  manner  just  d& 
scribed. 


^                   C           iV? 

B\                             7^ 

\                          t    K 

^^                     i    -^ 

^               7       ^ 

S                     L           ^ 

fa)       S,                _         -bl       ^ 

(a;     ^                           ^ 

^s^i.      /                           ^n 

I             ^K    r              3D 

— ,  M 

■^M 

.           .^   V 

7J                          1            ^            A 

t                           t         ^          ,^ 

H             '^             H^4              ^     i- 

t               t       ^^         \ 

e:       t\           ^^^  \ 

L 

^^           ^S                                  .--^^ 

z             s           ^     ^^ 

T                \       J         I 

-       •.   ^     -        -     ^^ 

^        -     V      £ 

-  /  ^^tx 

.                              J 

5                     L 

S^           ^Z      S^      ^2 

^>,^-^^        _t  — 

M      '      1 

One  Inch 

1    r  n 

Fig.  24 

The  very  best  conception  of  area  is  that  which  results  by  imagining 
the  rulings  on  the  squared  paper  just  mentioned  to  be  made  finer  and 
finer,  so  that  the  estimated  area  becomes  more  and  more  nearly  the  cor- 
rect area.  The  ideal  or  exact  area  bears  the  same  relation  to  these  esti- 
mates that  the  drawings  made  by  human  beings  do  to  the  ideal  figures  in 
geometry.     (See  §  11.) 

In  the  same  way,  lengths  of  curved  lines  can  be  estimated  by  first  re- 
placing each  small  bit  (arc)  of  the  curved  line  by  a  straight  line  joining 
the  ends  of  the  arc,  and  then  taking  the  sum  of  the  lengths  of  all  these 
pieces  of  straight  lines.    The  smaller  the  arcs,  the  more  accurate  this  result 


24  INTRODUCTION  [§25 

EXERCISES 

1.  How  many  exterior  angles  has  a  triangle  ? 

2.  In  a  certain  triangle  ABC  the  interior  angle  at  A  is  49°. 
What  is  the  exterior  angle  at  the  same  vertex  ? 

3.  Draw  three  triangles  of  different  shapes,  and  then,  using 
the  protractor,  determine  the  sum  of  the  three  interior  angles 
for  each  triangle.     Are  the  three  sums  equal,  and  if  so,  to  what  ? 

4.  The  end  of  the  minute  hand  of  a  clock  always  travels  in 
a  circle.     Why  ? 

5.  Draw  on  a  piece  of  squared  paper  a  rectangle  ^  in.  wide 
by  1  in.  high.  Draw  its  diagonal.  Estimate  the  area  in  each 
triangle  into  which  the  rectangle  is  divided.  Are  the  areas 
of  the  two  triangles  equal  ? 

6.  Draw  on  squared  paper  a  triangle  with  one  right  angle, 
and  with  the  perpendicular  sides  1.5  in.  and  .8  in.  long,  re- 
spectively.    Estimate  its  area. 

7.  Draw  on  squared  paper  a  rectangle  whose  diagonal  is 
the  longest  side  of  the  triangle  mentioned  in  Ex.  6.  Find  its 
area. 

8.  Draw  on  squared  pa^^er  two  concentric  circles,  with  the 
radius  of  one  twice  that  of  the  other.     Estimate  their  areas. 

9.  Construct,  on  squared  paper,  a  triangle  whose  sides  are, 
respectively,  2  in.,  1.5  in.,  and  1.7  in.     Estimate  its  area. 

[The  area  can  also  be  found  by  dividing  the  triangle  into  two  triangles 
that  have  one  right  angle  in  each,  by  a  perpendicular  from  one  corner  to 
the  opposite  side,  and  then  completing  each  of  these  smaller  triangles  into 
rectangles,  as  in  Exs.  5  and  7.] 

10.  What  is  the  sum  of  the  four  angles  of  a  rectangle  ? 

11.  A  courtyard  is  25  yd.  long  by  15  yd.  wide.  Draw  a  plan 
of  it  on  squared  paper,  scale  10  yd.  to  the  inch.  What  area 
is  represented  by  one  of  the  ruled  squares  on  your  paper  ? 
Find  the  area  of  the  courtyard. 


§25] 


FUNDAMENTAL  IDEAS 


25 


MISCELLANEOUS    EXERCISES    FOR   PART   II 

1.  In  Fig.  19,  p.  17,  the  angles  a  and  2/ are  vertical  angles.  If 
Z.X  =  40°,  what  is  the  value  of  Z  w,  its  supplement  ?  Since 
Z  w  is  also  the  supplement  of  Z?/,  what  is  the  value  oi  Z.y? 
Compare  Z  x  and  Z  y. 

2.  If  Z  a;  in  Fig.  19  is  38°,  what  is  the  value  oiy?  If  Z  ?/ 
is  78°,  what  is  the  value  of  Za;? 

3.  Name  pairs  of  vertical  angles 
in  the  adjoining  figure.  What  is 
the  value  ofZic-h  Zy  -\-  /.z?  ~ — '^^^ — ^ ^ 

4.  Draw  a  diagram  showing  the 
complement  and  the  supplement  of  ^ 
an  acute  angle  ABC.     What  is  their  difference  ? 

5.  Show  that  the  bisectors  of  two  adjacent  supplementary 
angles  are  perpendicular  to  each  other. 

6.  By  use  of  squared  paper  deter- 
mine (approximately)  the  number  of 
square  inches  in  the  adjoining  figure, 
taking  AB  —  4  in.,  BC  =  3  in.,  and 
(7^  =  2  in. 

7.  The  figures  (a),  (6),  (c)  of  Fig.  24, 
p.  23,  have  the  same  area.    Is  the  length 
of  the  boundary  therefore  the  same  for  each  of  them  ?     Esti- 
mate these  lengths. 

8.  In  what  ratio  is  the  drawing  of  a  house  reduced  from  the 
actual  house  if  a  distance  of  12  ft.  is  represented  by  a  line 
1^  in.  long  ?  What  actual  distance  is  represented  on  the  same 
drawing  by  a  line  2  in.  long  ? 

9.  Find  the  difference  in  longitude  at  two  places  on  the 
earth  if  the  difference  in  sun  time  is  2  hr.  30  min. 

10.  Find  the  difference  in  the  sun  time  between  two  places 
that  differ  in  longitude  by  30° ;  between  two  places  that  differ 
in  longitude  by  20°. 


26  INTRODUCTION  [§  26 


PAET  III.     STATEMENTS  FOR   REFERENCE 

26.  Assumptions.  In  Chapters  I-Y,  which  we  are  about  to 
study  and  in  which  many  of  the  principles  thus  far  used  are 
more  carefully  considered,  we  shall  make  use  of  certain  self- 
evident  general  statements.  These  statements  are  those  upon 
which  Geometry  is  based.  They  are  divided  into  two  classes, 
known  respectively  as  Axioms  and  Postulates.  Axioms  refer  to 
quantities  in  general,  that  is,  without  special  regard  to  geom- 
etry; postulates  refer  especially  to  geometry.  The  following 
lists  (§§  27,  28)  contain  axioms  and  postulates  that  will  be  clear 
at  this  time. 

27.  Axioms. 

1.  If  equals  are  added  to  equals,  the  sums  are  equal.  Thus, 
if  a  =  6  and  c=  d,  then  a-\-c  =  b-{-d. 

2.  If  equals  are  subtracted  from  equals,  the  remainders  are 
equal.     Thus,  if  a  =  5  and  c  =  d,  then  a—c=h  —  d. 

3.  If  equals  are  multiplied  by  equals,  the  products  are  equal. 
Thus,  ii  a  —  b  and  c  =  d,  then  ac  =  bd. 

4.  If  equals  are  divided  by  equals,  the  quotients  are  equal. 

Thus,  if  a  =  6  and  c  =  d,  then  -  =  -  •     In  applying  this  axiom 

c      d 

it  is  supposed  that  c  and  d  are  not  equal  to  zero. 

5.  If  equals  are  added  to  unequals,  the  results  are  U7iequal 
and  in  the  same  order.  Thus,  if  a  =  6  and  c>  d,  then  a-\-  c  > 
b-hd. 

6.  If  equals  are  subtracted  from  unequals,  the  results  are  un- 
equal and  in  the  same  order.  Thus,  it  a  >  b  and  c  =  d,  then 
a  —  c>b  —  d. 

7.  If  unequals  are  added  to  unequals  in  the  same  sense,  the 
results  are  unequal  i7i  the  same  order.  Thus,  if  a  >  6  and  c  >  d, 
then  a-\-c>b-{-  d. 


§  29]  STATEMENTS  FOR  REFERENCE  27 

8.  If  unequals  are  subtracted  from  equals,  the  results  are  uiia 
equal  in  the  opposite  order.  Thus,  if  a  =  6  and  c  >  d,  then 
a  —  cKh  —  d. 

9.  Quantities  equal  to  the  same  quantity,  or  to  equal  quanti- 
ties, are  equal  to  each  other.  In  other  words,  a  quantity  may  he 
substituted  for  its  equal  at  any  time  in  any  expression. 

10.  The  ichole  of  a  quantity  is  greater  than  any  one  of  its  parts. 

11.  Tlie  whole  of  a  quantity  is  equal  to  the  sum  of  its  parts. 

28.  Postulates. 

1.  Only  one  straight  line  can  he  drawn  joining  two  given 
points. 

2.  A  straight  line  can  he  extended  indefinitely. 

3.  A  straight  line  is  the  shortest  curve  that  can  he  drawn  be- 
tween two  points. 

4.  A  circle  can  he  described  about  any  point  as  a  center  and 
with  a  radius  of  any  length. 

5.  A  figure  can  be  moved  unaltered  to  a  new  position. 

6.  All  straight  angles  are  equal.  Hence,  also,  all  right  angles 
are  equal,  for  a  right  angle  is  half  of  a  straight  angle.     §  17. 

29.  Names  of  Statements.  Aside  from  the  above  axioms  and 
postulates,  the  words  Theorem,  Problem,  Proposition,  and  Co7'- 
ollary  will  hereafter  be  used  in  the  following  special  senses : 

Theorem.  A  statement  of  a  fact  which  is  to  be,  or  has  been, 
proved  is  called  a  theorem. 

Problem.  A  statement  of  a  construction  (see  §  20)  which  is 
to  be  made  is  called  a  problem. 

Proposition.  Either  a  theorem  or  a  problem  is  known  as  a 
proposition. 

Corollary.  A  theorem  which  follows  immediately  as  a  conse- 
quence of  some  other  theorem  is  called  a  corollary  of  it. 


28  INTRODUCTION  f§  30 

30.  Summary  of  Construction  Problems.  In  this  Intro 
duction  we  have  shown  how  to  make  the  following  construc- 
tions (with  ruler  and  compasses  alone)  : 

1.  To  construct  a  triangle,  each  of  whose  sides  is  equal  to  a 
given  length.     §  3,  p.  2. 

2.  To  construct  a  triangle,  whose  three  sides  are,  respec- 
tively, equal  to  three  given  lengths.     §  4,  p.  3. 

3.  To  construct  a  perpendicular  to  a  given  straight  line  at 
a  given  point  in  that  line.     §  5,  p.  4. 

4.  To  construct  a  perpendicular  to  a  given  line  from  a 
given  point  not  on  that  line.     §  6,  p.  5. 

5.  To  construct,  at  a  given  point  in  a  given  line,  another 
line  that  makes  an  angle  equal  to  a  given  angle  with  the  given 
line.     §  7,  p.  6. 

6.  To  divide  a  portion  of  a  straight  line  into  two  equal 
parts.     (To  bisect  a  line.)     §  8,  p.  7. 

7.  To  divide  a  given  angle  into  two  equal  parts.  (To  bisect 
an  angle.)     §  9,  p.  8. 

31.  Facts  or  Theorems  now  Known.  We  have  also  either 
assumed  or  proved  the  following  geometrical  facts : 

1.  All  radii  of  the  same  circle  are  equal.  §  2,  p.  1 ;  and 
§  23,  p.  21. 

2.  Circles  whose  radii  are  equal  can  be  placed  upon  each 
other  so  that  their  centers  and  their  circumferences  coincide 
(lie  exactly  upon  each  other). 

3.  Equal  angles  may  be  placed  upon  each  other  so  that 
their  vertices  coincide  and  their  corresponding  sides  fall  along 
the  same  straight  lines.  This  is,  in  fact,  what  we  mean  by 
equal  angles. 

4.  Two  straight  lines  have  at  most  one  point  in  common. 
See  postulate  1,  p.  27. 

5.  Two  circles  have  at  most  two  points  in  common.     See  §  8. 


§  31]  STATEMENTS  FOR  REFERENCE  29 

6.  A  straight  line  and  a  circle  may  have  at  most  two  points 
in  common. 

7.  At  a  given  point  in  a  given  line  only  one  perpendicular 
can  be  drawn  to  that  line.     (A  consequence  of  Problem  3,  §  5.) 

8.  Complements  of  the  same  angle,  or  of  equal  angles,  are 
equal.     Ex.  9,  p.  18. 

9.  Supplements  of  the  same  angle,  or  of  equal  angles,  are 
equal.     Ex.  10,  p.  18. 

10.  Vertical  angles  are  equal.     §  19. 

11.  If  two  adjacent  angles  have  their  exterior  sides  in  a 
straight  line,  they  are  supplementary.     §  18. 

12.  If  two  adjacent  angles  are  supplementary,  they  have 
their  exterior  sides  in  a  straight  line.     See  Fig.  18. 

13.  If  each  of  two  figures  can  be  placed  upon  a  third  figure 
so  as  to  coincide  with  it,  they  can  be  placed  upon  each  other 
so  that  they  coincide. 

14.  Any  desired  angle  may  be  drawn,  and  any  angle  may  be 
measured,  by  the  use  of  a  protractor.  (But  the  use  of  this 
instrument  is  not  permitted  when  a  figure  is  to  be  constructed. 
See  §  20.) 

15.  A  perpendicular  to  a  given  line  through  any  given  point 
may  be  drawn  by  means  of  a  set  square  or  a  drawing  triangle. 
(But  the  use  of  these  instruments  is  not  permitted  when  a  figure 
is  to  be  constructed.     See  §  20.) 

16.  The  area  of  a  rectangle  (in  terms  of  a  unit  square)  is 
equal  to  the  product  of  its  width  and  its  height,  measured  in 
units  of  length  equal  to  one  side  of  the  unit  square. 

17.  The  area  of  any  given  figure  is  greater  than  the  area  of 
any  figure  that  is  drawn  completely  within  it. 

18.  The  areas  of  two  figures  are  equal  if  they  consist  of  cor- 
responding portions  that  can  be  made  to  coincide. 


30 


INTRODUCTION 


[§3] 


SUPPLEMENTARY  EXERCISES 

1.  Draw  (using  protractor)  an  angle  of  50°.  Construct 
(using  ruler  and  compasses)  its  complement.  Measure  your 
new  angle  with  the  protractor  and  see  if  it  has  the  proper  num- 
ber of  degrees. 

2.  Draw  an  acute  angle  and  then  an  obtuse  angle.  In 
each  case  estimate  as  nearly  as  you  can  without  using  the  pro- 
tractor how  many  degrees  there  are  in  the  angle.  Then,  check 
your  estimate  by  measurement,  note  your  errors,  and  find  what 
fraction  of  the  correct  amounts  each  of  these  errors  is. 

3.  Find  the  angle  whose  complement  and  supplement  are  in 
the  ratio  4 :  13. 

4.  Show  how  to  construct  an  angle  of  45°;  an  angle  of 
22°  30'. 

5.  Construct  two  lines  that  bisect  (divide  into  two  equal 
parts)  each  other  at  right  angles. 

6.  Draw  the  patterns  shown  below.  Your  drawings  should 
be  twice  the  size  of  the  copies.  The  curves  are  formed  by 
joining  arcs  of  certain  circles. 


7.  A  traveler  wishes  to  go  due  north  but  finds  his  way 
barred  by  a  swamp.  He  therefore  goes  five  miles  northeast, 
then  five  miles  north,  then  five  miles  northwest,  and  he  now 
finds  himself  due  North  of  his  starting  point.  Draw  a  map 
(scale  one  mile  to  the  inch)  and  determine  by  measurement 
how  many  miles  he  lost  by  going  out  of  his  course. 


§31] 


SUPPLEMENTARY  EXERCISES 


31 


LIGHT 


8.  A  tower  is  observed  from  a  point  500  ft.  distant  from 
its  foot,  and  the  angle  the  line  of  sight  makes  with  the  hori- 
zontal is  found  to  be  15°.     What  is  the  height  of  the  tower  ? 

9.  It  is  a  principle  of  physics  that  when  a  ray  of  light  is  re- 
flected from  a  mirror,  the  re- 
flected  ray   makes   the  same 
angle  with  a  line  perpendicu- 
lar to  the  mirror  that  the  origi-    ^>^^^^%    ^    ^^    ^^^ 
nal  ray  makes  with  the  same 

line.     Show  that  the  original  ray  and  the  reflected  ray  also  make 
equal  angles  with  the  mirror  itself. 

10.  Two  forts  defending  the  mouth  of  a  river,  one  on  each 
side,  are  10  mi.  apart.  Their  guns  have  a  range  (possible 
shooting  distance)  of  4|  mi.  Draw  a  plan  (scale  1  mile  to 
the  inch)  showing  what  part  of  the  river  is  exposed  to  fire 
from  the  two  forts.  [Godfrey  and  Siddons.] 

11.  Construct  two  triangles  each  with  the  ^ 
sides  a,  b,  and  c,  as  indicated  in  the  adjacent 

figure.     See  §  4,  p.  3.     Cut  them  out  and  place        ^ ^ 

one  on  the  other  so  that  they  coincide.     What      C  a         B 

conclusions  do  you  draw  concerning  such  triangles  ? 

12.  Construct  two  triangles  each  with  the  base  AB  =  2  in., 
angle  ABC  =  angle  DEF,  and  angle  ACB  =  angle  DFE.  Cut 
one  of  these  triangles  from  the  paper  and  place  it  upon  the 
other  so  that  the  corresponding  parts  coincide.  What  conclu- 
sions do  you  draw  concerning  such  triangles  ? 


2  IN. 


2  IN. 


13.  The  length  of  a  rectangular  field  is  50  yd.  and  the 
length  of  the  entire  boundary  is  180  yd.  What  is  its  breadth 
and  its  area  ? 


32 


INTRODUCTION 


l§31 


14.  Construct  two  equal  angles,  ABC  and  DEF.  On  the 
sides  of  these  angles  lay  off  the  distances  BA  and  ED,  each 
2   in. ;    also   the   distances  BC  and   EF,  each  IJ  in.      Join 


MAKE      2  IN. 


MAKE       2  IN. 


AC  and  DF,  thus   forming   two   triangles   ABC  and   D^i^. 
What  conclusions  do  you  draw  concerning  such  triangles  ? 

15.  Draw  a  triangle  with  two  sides  AC  =  3  in.  and  AB 
=  5  in.  and  their  included  angle  =  35°.  (Use  protractor.) 
Draw  another  triangle  in  which  AC  =  3  in.,  AB  =  5  in., 
but  their  included  angle  =  20°.  Do  two  sides  alone  fix  (deter- 
mine) a  triangle  ? 

16.  Is  a  triangle  determined  if  two 
sides  and  their  included  angle  are  given  ?  DF^^ 
Show  the  relation  of  the  last  question 
to  the  following  fact  from  everyday  life : 
Two  pieces  of  board  AB  and  BC  hinged 
at  B  can  be  held  rigid  by  nailing  a 
crosspiece  DE  to  both  sides. 

17.  Many  designs  may  be  made  by  emphasizing  a  part  of  the 
lines  on  squared  paper,  or  the  diagonals  of  those  squares. 
Copy  and  complete  the  following;  and  also  invent  others. 


§31] 


SUPPLEMENTARY  EXERCISES 


33 


18.  To  construct  the  plan  for  a  Gothic 
window,  proceed  as  follows :  Take  any  line 
AB  and  divide  it  into  four  equal  parts. 
With  A  and  B  as  centers  and  a  radius  equal 
to  AB,  draw  arcs  intersecting  at  C.  What 
radius  and  what  centers  are  used  to  de- 
scribe the  small  arcs  ?  0  is  found  by  tak- 
ing A  and  B  as  centers  and  AK  as  a  radius.  Complete  the 
figure. 

19.  Construct  (by  ruler  and  compasses)  patterns  like  the 
following : 


<^ 

^ 

(b) 


(e) 


SYMBOLS   AND  ABBREVIATIONS 

The  following  symbols  and  abbreviations  will  be  used  for  the 
sake  of  brevity  throughout  the  present  book : 


=  equal,  or  is  equal  to 

Ax.      Axiom 

Tt  not  equal,  or  is  not  equal  to 

Cons.  Construction,  or  by  con- 

> greater  than 

struction 

<  less  than 

Cor.     Corollary 

^  is  congruent  to 

Def.     Definition 

_L  perpendicular,  or  is  perpen- 

Hyp.  Hypothesis,  or  by  hy 

dicular  to 

pothesis 

II    parallel,  or  is  parallel  to 

Iden.  being  identical 

~  similar,  or  is  similar  to 

Prop.  Proposition 

Z  angle 

rt.        right 

A  angles 

st.        straight 

A  triangle 

Th.      Theorem 

A  triangles 

Prob.  Problem 

O  parallelogram 

Fig.     Figure  or  diagram 

UJ  parallelograms 

....    and  so  on 

O  circle 

hence  or  therefore 

CD  circles 

^^  arc 

The  signs  +,  — ,  X,  -5-,  are  used  with  the  same  meanings  as 
in  algebra.     The  following  agreements  are  also  made : 

a  -5-  &  =  a/b  =  a:b 


CHAPTER   I 


RECTILINEAR  FIGURES 


PART  I.     TRIANGLES 

32.   Definitions.     It   is  desirable   to   distinguish    between 
several  kinds  of  triangles  as  follows: 


c  c 

Equilateral  Triangle    Isosceles  Triangle 

Fig.  25 


Scalene  Triangle 


An  equilateral  triangle  has  all  three  of  its  sides  equal. 
A  triangle  that  has  any  two  of  its  sides  equal  to  each  other 
is  called  an  isosceles  triangle. 

A  scalene  triangle  is  one  that  has  no  two  of  its  sides  equal. 

fC 


C  "  c 

Equiangular  Triangle    Acute  Trla.ngle 

Fig.  26 


Obtuse  Triangle 


An  equiangular  triangle  has  all  three  of  its  angles  equal. 
An  acute  triangle  is  one  whose  angles  are  all  acute. 
An  obtuse  triangle  is  a  triangle  that  has  one  obtuse  angle. 

36 


36 


RECTILINEAR  FIGURES 


[I,  §  32 


A  triangle  one  of  whose  angles  is  a  right  angle  is  called  a 
right  triangle.  In  such  a  triangle,  the  side  opposite  the  right 
angle  is  called  the  hypotenuse,  and 
the  word  "side"  is  used  only  for 
the  other  two  sides,  as  indicated  in 
Fig.  27. 

The  side  upon  which  any  triangle 
appears  to  rest  is  called  its  base. 

The  vertex  opposite  to  the  base  is 
called  the  vertex  of  the  triangle,  and  the  angle  opposite  to  the 
base  is  called  the  angle  at  the  vertex. 

The  perpendicular  distance  from  any  vertex  to  the  opposite 


Right  Triangle 
Fig.  27 


c      M      P      B 


c   M     B 


.      Fig.  28 

side  (extended  if  necessary)  is  called  an  altitude  of  the  tri- 
angle. 

The  distance  from  any  vertex  to  the  middle  point  of  the 
opposite  side  is  called  a  median  of  the  triangle. 

Any  portion  of  a  straight  line  between  two  points  is  called 
a  segment  of  that  line.  Thus  the  sides,  altitudes,  and  medians 
of  any  triangle  are  line  segments. 

Any  figure  composed  wholly  of  points  and  straight  lines  is 
called  a  rectilinear  figure. 

33.  Congruent  Figures.  Two  triangles  that  can  be  made  to 
fit  each  other  exactly  (coincide)  by  properly  placing  the  one 
upon  the  other  are  called  congruent.  More  generally,  any  two 
geometric  figures  are  congruent  if  they  can  be  made  to  coincide 
exactly. 


I,  §  34] 


TRIANGLES 


37 


When  two  congruent  figures  are  made  to  coincide,  any  cor- 
responding parts  coincide :  corresponding  angles  of  congruent 
figures  are  equal ;  corresponding  lengths  are  equal;  any  portion 
of  one  of  two  congru- 
ent figures  is  congru-      ^       ^ P 

ent  to  the  correspond-         ^       "     ' 

ing    portion    of    the  ^       \    3    ^^ 

other. 

An   illustration  of 
two     congruent     fig- 
ures, each  broken  up  into  certain  corresponding  parts  which 
are  also  congruent,  is  given  in  Fig.  29. 


Fig 


EXERCISES 

1.  How  many  altitudes  has  a  triangle?  How  many  me- 
dians ?     Draw  figures  illustrating  your  answers. 

2.  Can  a  right  triangle  be  isosceles  ?     Draw  figures. 

3.  Are  all  right  triangles  isosceles  ?      Draw  figures. 

4.  Can  a  right  triangle  be  equilateral?  Explain  your 
answer. 

5.  What  kind  of  triangle  was  drawn  in  Problem  1,  p.  2? 
In  Problem  2,  p.  3  ? 

6.  Investigate  the  following  questions :  (1)  Is  there  any 
kind  of  triangle  for  which  the  medians  coincide  with  the  alti- 
tudes, and  also  with  the  bisectors  of  the  three  angles  ?  If  so, 
describe  it.  (2)  Will  these  lines  usually  all  be  different  for  a 
triangle  ?     Illustrate  your  answer  by  drawings. 

34.  Congruence  of  Triangles.  We  now  proceed  to  state  and 
prove  certain  theorems  regarding  triangles.  The  pupil  should 
first  reread  carefully  §  22. 

Our  first  topic  of  study  will  be  the  following  question: 
"  When  are  two  triangles  congruent  ?  "  Note  again  the  defini- 
tion of  congruent  figures  as  given  in  §  33. 


38  RECTILINEAR  FIGURES  [I,  §  35 

35.  Theorem  I.  If  two  triangles  have  two  sides  and  the 
included  angle  of  the  one  equal,  respectively,  to  two  sides  and 
the  included  angle  of  the  other,  the  triangles  are  congruent. 


Fig.  30 

In  the  figure,  let  ABC  and  A'B'C  be  the  two  triangles,  and 
let  us  suppose  that  we  know  (as  the  theorem  says)  that  AB  = 
A'B',  AC=  A' a,  and  that  the  angle  A  =  the  angle  A\ 

We  are  now  to  prove  that  these  two  triangles  are  congruent, 
which  means  that  we  have  to  show  that  the  one  may  be  fitted 
on  to  the  other  so  that  they  will  exactly  coincide  in  all  their 
parts. 

Kow,  since  AB=:A'B',  we  can  place  the  triangle  A'B'C  on 
the  triangle  ABO  so  that  A'B'  will  coincide  with  its  equal 
AB,  making  the  point  C"  fall  somewhere  on  the  same  side  of 
AB  as  a 

Then,  A'C  will  lie  along  AO,  because  angle  A'  =  angle  A, 
this  being  also  one  of  the  given  (supposed)  facts. 

Moreover,  C  will  fall  exactly  at  (7,  since  A'C  =  AC,  which 
is  another  of  the  given  facts. 

It  thus  follows  that  the  side  C'B'  will  fit  exactly  upon  CB, 
for,  according  to  Postulate  I,  only  one  straight  line  can  be 
drawn  through  the  two  points,  C,  B. 

Therefore,  the  two  triangles  are  congruent. 

Note.  The  fact  stated  in  this  theorem  was  indicated  in 
Exs.  2  and  3,  p.  6 ;  and  in  Ex.  14,  p.  32.  In  those  exercises, 
however,  the  truth  was  only  suggested.  What  we  do  in  a  proof 
such  as  that  just  given  is  to  make  certain  what  was  previously 
simply  plausible. 


I,  §  36]  TRIANGLES  39 

36.    Corollary  I.     Two  right  triangles  are  congruent  if  the  two 
sides  of  the  one  are  equal  respectively  to  the  two  sides  of  the  other. 

Reasoning  from  Theorem  I,  the  student  should  convince 
himself  of  the  truth  of  this  corollary. 

[Hint.  First,  note  the  form  of  all  right  triangles,  as  illustrated  in 
Fig.  27.  Note  also  from  the  same  figure  how  the  word  "side  "  is  used 
for  such  triangles.  Now  draw  two  such  triangles  having  the  "sides"  of 
the  one  equal  respectively  to  the  "  sides  "  of  the  other,  and  see  how  Theo- 
rem I  applies  to  them.] 

EXERCISES 

"^1.   Write  out  the  proof  of   Theorem  I  for  two  triangles 
shaped    as    in   the    adjoining    figure,   in   which   we   suppose 
AB^A'B',    AC  =  A'C,    and 
angle  A  =  angle  A'. 

2.  Using  a  protractor,  draw 

a  triangle  in  which  one  angle  is     A"*^-- 'B     A-^ 'b' 

30°  and  the  two  sides  that  in- 
clude that  angle  are  respectively  3  and  4  inches  long.     Show, 
by  Theorem  I,  that  if  another  triangle  is  drawn  which  has 
these  parts,  it  is  congruent  to  the  one  first  drawn. 

3.  In  the  figure,  L  represents  a  lake.  It  is  required  to  find 
its  length,  i.e.  the  distance  between  A  and  B.  Show  that  this 
may  be  done  as  follows :  .^^^^^^^ 

(1)  Fix  a  stake  at  some  convenient  point  ^^^       ^^ 
C,  and  measure  the  distances  AO  and  BC.  \y^ 

(2)  In  line  with.^C  set  a  stake  D,  such  /C\^ 
that  CD  =  AC.     Then  in  line  with  BC,  set  a  /        \ 
stake  E,  such  that  CE  =  BC.  E-- -^D 

The  distance  from  stake  D  to  stake  E  will  be  the  required 
distance  between  A  and  B.  Show  that  this  is  a  consequence  of 
Theorem  I. 

4.  Show  how  the  method  of  Ex.  3  could  be  applied  to  de- 
termine the  greatest  length  of  your  school  building. 


40  RECTILINEAR  FIGURES  [I,  §  37 

37.  Theorem  II.  If  two  triangles  have  two  angles  and  the 
included  side  in  the  one  equal,  respectively,  to  two  angles  and  the 
included  side  in  the  other,  the  triangles  are  congruent. 


In  the  figure,  let  ABC  and  A'B'C  be  the  two  triangles,  and 
let  us  suppose  that  we  know  that 

^A  =  ZA',ZB  =  ZB',^dAB  =  A'B'. 

[For  brevity  we  here  use  the  symbol  Z  for  the  word  angle  (see  p.  34). 
Also,  mstfad  of  "  side  AB  =  side  A'^  "  we  write  simply  AB  =  A'B'.^ 

We  shall  now  prove  that  the  triangles  ABC  and  A'B'C  are 
congruent  by  showing  that  it  is  possible  to  place  the  one  upon 
the  other  so  that  they  shall  coincide. 

To  carry  out  the  proof,  place  the  triangle  A'B'Cf  upon  the 
triangle  ABC  so  that  A'&  coincides  with  its  equal  AB,  and  so 
that  O  and  C  fall  upon  the  same  side  of  AB. 

Then  A'C  will  fall  along  AC  because  ZA  =  Z  A',  which 
was  one  of  the  given  facts. 

Moreover,  B'C  wiU  fall  along  BC,  since  Z  B  =  Z  B',  which 
was  also  given. 

It  follows  from  this  that  C  will  falT  exactly  at  C,  for, 
by  Statement  4,  §  31,  the  two  lines  AC  and  BC c&n  intersect  in 
only  one  point. 

Thus,  the  two  triangles  can  be  made  to  coincide  completely, 
and  they  are,  therefore,  congruent.  Using  the  symbols  of 
p.  34,  this  result  is  written  in  the  following  form : 

A  ABC  ^  A  A'B'C 


I,  §  39] 


TRIANGLES 


41 


38.  Corollary  1.  Two  right  triangles  are  congruent  if  an  acute 
angle  and  its  adjacent  side  in  one  are  equal,  respectively,  to  an  acute 
angle  and  its  adjacent  side  in 
the  other. 

Thus,   the   two   right   tri- 
angles ABC  and  A'B'C  are 
congruent  if  AB  =  A'B'  and 
Z.A  =  Z.A'.     This  corollary,  like  all  others,  should  be  proved 
by  the  student. 

EXERCISES 

1.  Draw  a  right  triangle  having  one  angle  30°  and  the  adja- 
cent side  (not  the  hypotenuse)  3  in.  long.  Measure  the  other 
angle  with  a  protractor,  and  measure  each  of  the  other  sides. 
Repeat  this  with  an  angle  of  45°  in  place 
of  30°. 

2.  Theorem  II  was  used  by  Thales 
(640-546  B.C.)  to  determine  the  distance 
of  a  vessel  V  from  the  shore,  by  measur- 
ing the  angles  w  and  v  and  then  con- 
structing the  congruent  triangle  ABX  on  shore, 
this  can  be  done. 

3.  A  drawing  triangle  is  usually 
made  of  celluloid,  with  one  angle  (Z  C 
in  the  figure),  a  right  angle.  One  other 
angle  A  is  usually  made  30°.  Show  that 
if  the  length  AC  is  chosen,  the  form  of 
the  triangle  is  completely  determined. 


Explain  how 


b  C 

Drawing  Triangle 


39.  Form  of  Proofs.  Hereafter  we  shall  use  symbols  when- 
ever possible,  thus  enabling  us  to  give  proofs  in  a  more  con- 
densed form.     See  the  list  of  symbols,  p.  34. 

Xote  that  in  the  arrangement  of  the  proof  the  different  steps 
are  in  the  left  column,  while  the  reason  for  each  step  is  in  the 
right  column. 


42  RECTILINEAR  FIGURES  [I,  §  40 

40.  Theorem  III.     In  an  isosceles  triangle  the  angles 
opposite  the  equal  sides  are  equal. 


Given  the  isosceles  triangle  ABC  in  which  AC  —  BO, 
To  prove  that  ZA=ZB. 

Proof.     Draw  CD  bisecting  Z  ACB. 
Then,  in  the  A  ACD  and  BCD  we  have 

CD  =  CD,  Men. 

AC=:BC,  Given 

ZACD  =  ZBCD.  Cons. 

Therefore  A  ACD  ^  A  BCD ;       .  §  35 

and  hence  ZA  =  ZB.  §  33 

41.  Corollary  1.  If  a  triangle  is  equilateral.,  it  is  also  equi- 
angular. 

State  carefully  the  reasons  for  this  conclusion,  in  the  form 
of  the  proof  of  Theorem  III. 

EXERCISES 

1.  Construct  (using  ruler  and  compasses)  an  isosceles  triangle 
and  measure  by  means  of  the  protractor  the  angles  opposite 
the  equal  sides.  Do  your  results  conform  to  what  Theorem 
III  says  ? 


I,  §  40] 


TRIANGLES 


43 


2.  An  ordinary  gable  roof  has  a  form  (cross  section)  such 
as  indicated  in  the  accompanying  figure.  How  does  this  illus- 
trate Theorem  III  ? 

Ridge 


3.  Mention  some  other  familiar  object  in  which  an  isosceles 
triangle  occurs,  draw  a  figure  to  represent  it,  and  state  how 
it  illustrates  Theorem  III. 

4.  Prove  that  the  medians  drawn  through  the 
base  angles  of  an  isosceles  triangle  are  equal.  Write 
out  the  proof  in  the  precise  form  used  in  the  proof 
of  Theorem  III,  using  two  columns,  one  for  state- 
ments of  fact  and  the  other  for  the  reasons ;  and  use 
the  symbols  of  p.  34. 

[Hint.  Given  the  isosceles  A  ABC  in  which  AC  =  BC  and  let  AD 
and  BE  be  the  medians  drawn  through  the  base  angles  A  and  B. 
To  prove  AD  =  BE.  In  the  proof,  direct  your  attention  to  the  k^EAB 
and  DBA.  They  have  Z  EAB  =  Z  DBA  by  Theorem  III ;  at  the  same 
time  they  have  AE  =  BD  because  each  is  one  half  of  the  equal  sides  AG 
and  BC.  Also  AB  =  AB.  From  these  facts,  reason  by  means  of 
Theorem  I  that  A  ABE  ^  A  ABD,  and  hence  AD  =  EB,  which  was  to 
be  proved.] 

5.  Prove  that  the  bisectors  of  the  base  angles  of  an  isosceles 
triangle  are  equal. 

[Hint.  Draw  a  figure  similar  to  that  of  Ex.  4,  but  make  AD  bisect 
Z  A,  and  BE  bisect  Z  B.  Direct  your  attention  to  the  A  EAB  and 
DBA.  They  have  Z  EAB  =  Z  ABD  by  Theorem  in  ;  at  the  same  time 
Z  EBA  =  Z  DAB,  being  halves  of  equal  angles.  Whence,  show  by 
Theorem  II  that  A  EAB  ^  A  DBA.  Then,  AD  =  BE,  as  was  to  be 
proved.] 


44  RECTILINEAR  FIGURES  [I,  §  42 

42.  Nature  of  Proofs  in  Geometry.  Hypothesis.  Conclu- 
sion. Every  proof,  as  arranged  in  the  form  which  we  have 
now  adopted  and  as  illustrated  by  the  proof  of  Theorem  III, 
consists  of  four  parts,  as  follows : 

(1)  A  careful  statement  of  the  theorem,  accompanied  by 
an  appropriate  figure. 

(2)  A  condensed  statement  of  what  we  have  to  work  with, 
or  what  is  given.  This  part  of  the  statement  of  the  theorem 
is  called  its  hypothesis,  or  thing  supposed.  Thus,  in  Theorem 
III  the  hypothesis  is  that  the  triangle  is  isosceles. 

(3)  A  condensed  statement  of  what  is  to  be  proved.  This 
is  called  the  conclusion. 

Thus,  in  Theorem  III  the  conclusion  is  that  "the  angles 
opposite  the  equal  sides  are  equal." 

(4)  The  details  of  the  proof.  This  consists  of  a  series  of 
statements,  each  accompanied  with  its  reason. 

EXERCISES 

1.  What  is  the  hypothesis  of  Theorem  I?  What  is  its 
conclusion  ?     Answer  the  same  questions  for  Theorem  II. 

2.  Hypotheses  and  conclusions  occur  in  all  careful  argu- 
ments in  other  subjects  as  well  as  in  Geometry.  Pick  out  the 
hypothesis  and  the  conclusion  in  each  of  the  following  state- 
ments : 

(a)  If  he  is  guilty,  he  should  be  punished. 

(5)  If  a  piece  of  iron  is  magnetized,  it  will  attract  other 
pieces  of  iron. 

(c)  A  body  heavier  than  water  will  sink  in  water. 

(d)  Corollaries  of  §§  36,  38,  41. 

3.  Make  some  conditional  statement  similar  to  those  of 
Ex.  2,  and  give  the  hypothesis  and  the  conclusion. 

4.  Make  some  statement  that  you  think  is  true  about  some 
geometric  figure.  Pick  out  the  hypothesis  and  the  conclusion. 
Draw  a  figure  to  illustrate  the  statement. 


I,  §  44] 


TRIANGLES 


45 


43.  Theorem  IV.  The  hisectoi'  of  the  angle  at  the 
vertex  of  an  isosceles  triangle  is  perpendicular  to  the 
base  and  bisects  the  base. 


isosceles    A  ABC  in 
bisects    the    vertical 


Given  the 
which    CD 

za 

To  prove  that  CD  ±  AB  and  that 
AD  =  DB. 

Proof.     In  the  A  ADC  and  DBC 
we  have 

AG=  CB,  Given 

and        Z  ACD  =  Z  DCS.     Given 
Moreover  CD  =  CD. 

Therefore  A  ADC  ^  A  DBC ; 

whence  Z  ADC  =  Z  BDC, 

so  that  CD±AB. 

Also  AD  =  DB. 


§  17 
§33 


44.  Corollary  1.  In  any  isosceles  triangle  (a)  The  bisector  of 
the  angle  at  the  vertex  divides  the  triangle  into  two  congruent 
right  triangles. 

(h)  The  bisector  of  the  vertical  angle  coincides  with  both  the 
altitude  and  the  median  drawn  through  the  vertex. 

(c)  The  perpendicular  bisector  of  the  base  passes  through  the 
vertex^  and  divides  the  triangle  into  two  congruent  right  triangles. 

EXERCISES 

1.  If  a  plumb  line  (a  string  with  a  heavy  bob  attached  at 
one  end)  be  let  down  from  the  ridge  of  the  roof  represented 
in  Ex.  2,  p.  43,  the  bob  strikes  the  floor  in  a  way  that  illustrates 
Theorem  IV.     How  ? 

2.  Prove  that  if  the  bisector  of  the  angle  at  the  vertex  of  any 
triangle  is  perpendicular  to  the  base,  the  triangle  is  isosceles. 

[HiKT.  Apply  Theorem  II  to  the  &.ADC,  BDC  (Fig.  33),  and  show 
that^C=  CB.] 


46  RECTILINEAR  FIGURES  fl,  §  45 

45.  Theorem  V.  If  two  triangles  have  three  sides  of  the  one 
equal  respectively  to  the  three  sides  of  the  other,  they  are  con- 
gruent 


c 


Fig.  34 

Given  the  j^ABG  and  A^B'O  in  which  AB  =  A^B\  BO  = 
B'C,  and  CA  =  C'A', 

To  prove  that  A  ABC  ^  A  A' B'C. 

Proof.  Place  AA'B'G'  in  the  position  ABG",  thus  making 
the  side  A'B'  coincide  with  its  equal  AB  and  causing  the  point 
C  to  take  up  the  position  marked  C". 

Join  G  and  C"  by  a  straight  line.  Then,  in  AAGG",  we 
have  AG=AG".  Given 

Therefore  Z  AGG"=Z  AG"G.  §  40 

Likewise,  in  A  BGG",  we  have 

BG=BG";  Given 

hence  Z  BGG" = Z  BG" G.  Why  ? 

Therefore  ZAGG"-\-ZBGC  =  ZAG"G-\-ZBG"G,  Ax.  1  (§  27) 
that  is,  ZAGB=ZACB; 

whence  A  AGB^A  AG"B,  §  35 

that  is,  A  ABG  ^  A  A' B'C. 

46.  Corollary  1.  Three  sides  determine  a  triangle;  that  is, 
if  the  three  sides  are  given,  the  triangle  is  thereby  fixed. 

The  statement  means  that  if  the  three  sides  of  a  triangle 
are  known,  any  triangle  made  with  these  sides  is  congruent  to 
any  other  triangle  that  has  the  same  sides.  See  Problem  2, 
p.  3 ;  Exs.  2,  3,  p.  3 ;  Ex.  11,  p.  31. 


I,  §  46]  TRIANGLES  47 

EXERCISES 

1.  If  three  boards  are  nailed  together  in  the  form  of  a  tri- 
angle, with  one  nail  at  each  corner,  will  the  frame  thus  made 
be  quite  stiff?  Show  how  the  corollary  just  stated  is  related 
to  your  answer. 

2.  Two  circles  whose  centers  are  at 
0  and  0'  intersect  in  A  and  B.  Prove 
thsit  A  OAO'  ^  A  OBO'. 

3.  Prove  that  two  triangles  ABC  and 
A'B'C    are    congruent   if    AB  =  A'B', 

BC  =  B'C,   and  the  median  through  A  equals   the   median 
through  A'. 

[Hint.  Draw  the  two  triangles.  Let  D  be  the  point  of  intersection 
of  the  median  through  A  with  the  side  ^C  in  A  ABC,  and  D'  the  cor- 
responding point  in  A  A' B'C.  First  show,  by  Theorem  V,  that  AADB 
^  A  A'D'B',  remembering  that  BD  =  ^  BC  and  B'D'  =  \  B'D'.  Hence 
show  that  ZADC  =  ZA'D'C'  by  using  9,  p.  29.  Finally,  show  that 
AADCc^AA'D'C.    Then  AABC^AA'B'C  by  §  33.] 

4.  The  framework  of  bridges,  scaffolding,  and  other  struc- 
tures consists  usually  of  a  network  of  triangles  whose  sides  are 
stiff  pieces  of  iron  or  wood.  Show,  by  means  of  Theorem  V 
and  Corollary  1,  why  the  entire  structure  is  stiff. 

5.  Show  that  the  temporary  bracing  of  a  window  frame  in 
a  building  during  its  erection  by  a  board  nailed  to  the  frame 
and  to  the  floor,  is  an  illustration  of  §  46. 

6.  Point  out  the  triangle  in  the  roof  construction  of  Ex.  2, 
p.  43,  which  makes  the  roof  structure  rigid. 

Note.  These  Exercises  illustrate  the  great  importance  of  Theorem  V 
in  structures  of  all  kinds.  The  practical  value  of  the  study  of  triangles 
arises  principally  from  the  frequent  application  of  Theorems  I,  II,  and  V, 
§§  35,  37,  45,  both  in  practical  affairs  and  in  the  proofs  of  other  geometric 
theorems.  These  three  theorems  are  printed  in  boldface  type  to  indicate 
their  especial  importance  ;  they  should  be  studied  most  carefully. 


48  RECTILINEAR  FIGURES  [I,  §47 

47.  Theorem  VI.  An  exterior  angle  of  a  triangle  is  greater 
than  either  of  the  opposite  interior  angles. 

Given  the  A  ABC  hav- 
ing the  exterior  Z  DBC.  ^ 

To  prove 

that  Z  DBC  >  A  A; 
alsothatZi>J5(7>Z(7. 

Proof.  Through  E,  the 
mid-point  of  BC,  draw 
AE  and  prolong  it  to  F, 
making  EF—  AE.  Also, 
draw  BF. 

Then,  in  the  A  AEC  and  BFE  we  have 

AE  =  EF,  and  BE  =  EC,  Cons, 

and  Z  CEA  =  Z  JS^i/'.  10,  §  31 

Therefore  A  AEC  ^  A  BFE ;  Why  ? 

whence  Z  O  =  Z  i^^J^.  Why  ? 

But  Z  i)5^  >  Z  i^^^;;  Ax.  10 

hence  Z  D5^  >  Z  C,  Ax.  9 

that  is,  Z.DBC  >  ZC,  which  was  to  be  proved. 

[Similarly,  by  bisecting  AB  and  proceeding  as  above  it  can  be  proved 
that  Z.  ABK,  which  equals  Z  DBC  (why  '?),  is  greater  than  Z  A.^ 

EXERCISES 

1.  By  drawing  a  series  of  triangles,  determine  what  form  a 
triangle  tends  to  take  as  one  of  its  exterior  angles  comes  closer 
and  closer  in  size  to  either  of  its  opposite  interior  angles. 

2.  Prove  that  no  triangle  can  have  two  right  angles. 

[Hint.  If  Z  O,  Fig.  35,  were  equal  to  90",  then  ZDBC  would  be 
greater  than  90°  (§47)  ;  hence  Z  ABC  <90°,  since  it  is  the  supplement 
of  Z  DBC.     Write  out  the  proof  in  full.  ] 

3.  Prove  that  no  triangle  can  have  two  obtuse  angles. 

4.  Prove  that  the  bisector  of  an  exterior  angle  of  a  triangle 
is  perpendicular  to  the  bisector  of  the  adjacent  interior  angle. 


I,  §50]  PARALLEL  LINES  49 

PAKT  II.    PARALLEL   LINES 

48.    Definitions.     Lines  that  lie  in  the  same  plane  but  do  not 
meet  however  far  they  may  be  extended  are  said  to  be  parallel 

A B 


Parallel  Lines 
Fig.  36 

If  two  lines  are  cut  by  a  third  line,  this  third  line  is  called 
a  transversal,  and  the  angles  at  the  points  of  intersection  are 
named  as  follows : 

tf  u,  X,  and  w  are  interior  angles. 

Sj  V  J  y,  and  z  are  exterior  angles. 

s  and  z,  or  v  and  y,  are  alternate  ex- 
terior angles. 

t  and  IV,  or  u  and  x,  are  alternate 
interior  angles. 

V  and  w,  or  s  and  x,  or  ti  and  z, 
or  t  and  y,  are  corresponding  angles. 

49.  Parallel  Postulate.  Besides  the  postulates  of  §^  28,  the 
following  is  necessary  for  the  study  of  parallel  lines : 

Parallel  Postulate.  Only  one  line  can  be  drawn  through  a 
given  point  parallel  to  a  given  line.  That  is,  any  other  line  so 
drawn  will  coincide  with  the  first  one. 

This  postulate  was  stated  in  a  somewhat  different  form  by 
Euclid  (about  300  b.c).  It  was  recognized  by  him  and  by 
later  mathematicians  that  it  is  peculiarly  important.  It  is  often 
called  the  Euclidean  postulate. 

50.  Corollary  1.  Lines  parallel  to  the  same  line  are  parallel 
to  each  other.  For,  if  two  lines  parallel  to  one  and  the  same  line 
should  meet,  there  would  be  more  than  one  parallel  through  a 
point,  thus  contradicting  the  preceding  postulate. 


50  RECTILINEAR  FIGURES  [I,  §51 

51.   Theorem  VII.     When  two  lines  are  cut  by  a  transversal, 
if  the  alternate  interior  angles  are  equal,  the  two  lines  are  parallel. 

T 


■— --.  K 


S  Fig.  38 

Given  any  two  lines  AB  and  CD  cut  by  the  transversal.  ^iST  at 
E  and  F,  so  that  the  alternate  interior  angles  x  and  y  are  equal. 

To  prove  that  AB  is  parallel  to  CD. 

Proof.  Let  us  suppose  for  the  moment  that  AB  and  CD  are 
not  parallel.  In  this  case  they  would  meet  in  some  point 
which  we  will  call  K,  and  a  triangle  EFK  would  be  formed. 
Then  the  exterior  angle  x  of  this  triangle  would  be  greater 
than  its  interior  angle  y,  by  Theorem  VI.  But  this  is  impossi- 
ble, since  our  hypothesis  is  that  Z.x  =  /.y.  In  other  words, 
the  conclusion  that  AB  and  CD  are  not  parallel  cannot  be  true 
if  our  hypothesis  that  /.x  =  Zyis  true.  The  only  other  possible 
conclusion  is  that  AB  and  CD  are  parallel. 

52.  Corollary  1.  Lines  perpendicular  to  the  same  line  are 
parallel. 

EXERCISES 

1.  If  several  strips  are  nailed  to  a  board  at  right  angles  to 
it,  show  that  the  strips  are  parallel. 

2.  If  another  board  is  nailed  perpendicular  to  the  strips  of 
Ex.  1,  show  that  it  is  parallel  to  the  first  board. 

3.  Draw  any  line  CD  (Fig.  38)  and  select  any  point  E  not 
on  it.  Draw  any  line  through  E  to  cut  CD  at  some  point,  F. 
Lay  off  /.x  =  Z-y  with  a  protractor,  or  as  in  §  7,  p.  6.  Show 
that  this  process  gives  a  line  parallel  to  CD  through  E. 


I,  §53]  PARALLEL  LINES  51 

53.  The  Indirect  form  of  Proof.  Argument  by  Reduction 
to  an  Absurdity.  The  student  may  have  observed  that  the 
proof  just  given  in  §  51  differs  essentially  from  other  proofs  we 
have  given.  In  the  proof  of  Theorem  VII  we  meet  for  the 
first  time  what  is  called  an  indirect  proof,  otherwise  known  as  a 
reductio  ad  absurdum,  or  reduction  to  an  absurdity.  This  form 
of  proof  is  of  frequent  occurrence.  In  substance,  an  indirect 
proof  of  a  theorem  consists  in  first  supposing  something  differ- 
ent from  the  theorem's  conclusion,  and  then  showing  that  an 
absurdity  results,  thus  leaving  the  theorem  itself  as  the  only 
possible  statement  of  fact. 

The  use  of  the  indirect  proof  is  common,  not  only  in  geom- 
etry, but  also  in  many  of  the  familiar  experiences  of  everyday 
life.  Suppose,  for  example,  that  on  a  certain  night  a  robbery 
is  committed  in  a  certain  store  and  the  next  day  Mr.  A  is  sus- 
pected of  having  done  it.  He  succeeds  in  proving,  however, 
that  throughout  the  night  in  question  he  was  in  another  town. 
This  is  sufficient  to  free  him  of  suspicion.  Why  ?  —  Because 
the  supposition  that  he  is  guilty  is  thus  made  to  lead  to  the 
absurd  conclusion  that  he  was  in  two  different  places  at  the 
same  time.  The  very  strongest  kind  of  argument  is  to  show 
that  some  contention  of  an  opponent  leads  to  absurd  conclusions. 

EXERCISES 

1.  A  closed  wooden  box  is  known  to  contain  a  piece  of 
metal.  A  magnet  is  brought  near  and  found  to  be  attracted. 
The  conclusion  is  drawn  that  the  metal  within  the  box  is  either 
iron  or  steel.  Show  that  this  conclusion  is  drawn  by  a  process 
of  indirect  proof. 

2.  Using  Theorem  III,  give  an  indirect  proof  of  the  follow- 
ing theorem :  If  no  two  of  the  three  angles  of  a  triangle  are 
equal,  the  triangle  cannot  be  isosceles. 

3.  Show  that  the  reasoning  used  in  §  50  is  really  a  reduction 
to  an  absurdity. 


52  RECTILINEAR  FIGURES  [I,  §54 

54.   Theorem  VIII.    If  two  parallel  lines  are  cut  by  a  trans< 
versal,  the  alternate  interior  angles  are  equal. 


V 

d___ 

A 

D 

c 

s^ 

/ 

A     

Fig.  39 

d' 

Given  the  two  parallel  lines  AB  and  CD  cut  by  the  trans- 
versal ST  Sit  E  and  jP  and  making  the  alternate  interior  angles 
X  and  y. 

To  prove  that  the  angles  x  and  y  are  equal. 
.  Proof.     Suppose  A  x  ^  /.y      (For  the  symbol  ^,  see  p.  34.) 

Now  draw  CD'  through  F,  making  Z  D'FT  =  Z.t. 

Then  CD'WAB',  §51 

but  CD  li  AB.  Given 

Thus  we  should  have  two  different  lines  through  the  same 
point  F  parallel  to  the  line  AB,  which  is  impossible,  according 
to  the  postulate  of  §  49. 

Therefore  Z  x  =  Zy,  which  was  to  be  proved. 

55.  The  Converse  of  a  Theorem.  The  student  may  have 
observed  that  Theorems  YII  and  VIII,  though  not  the  same,  are 
very  closely  related.  Careful  examination  of  the  two  will  show 
that  the  precise  nature  of  this  connection  lies  in  the  fact  that 
the  hypothesis  and  conclusion  of  the  one  have  become,  respec- 
tively, the  conclusion  and  hypothesis  of  the  other.  In  other 
words,  the  one  is  the  other  simply  turned  about.  In  general, 
when  any  two  theorems  are  related  in  this  way,  the  one  is  said 
to  be  the  converse  of  the  other. 

Aside  from  geometry,  there  are  many  instances  of  state- 
ments and  their  converse.     The  following  will  illustrate  this 


I,  §55]  PARALLEL   LINES  53 

fact,  and  also  the  fact  that  because  a  theorem  or  other  state- 
ment is  true  we  cannot  always  be  certain  that  its  converse 
is  true. 

{Statement)     If  a  boat  is  made  of  wood,  it  floats.     {True) 
{Converse)      If  a  boat  floats,  it  is  made  of  wood.     {False) 
An  instance  in  Geometry  in  which  the  converse  of  a  true 
statement  is  false  is  : 

{Statement)    If  a  figure  is  a  square,  it  is  a  rectangle.    {True) 
{Converse)     If  a  figure  is  a  rectangle,  it  is  a  square.    {False) 
An  instance  in  Geometry  in  which  both  the  original  and  the 
converse  theorems  are  true  is : 

{Statement)  If  two  sides  of  a  triangle  are  equal,  the  angles 
opposite  them  are  equal.     Theorem  III,  §  40. 

{Converse)  If  two  angles  of  a  triangle  are  equal,  the  sides 
opposite  them  are  equal.     Theorem  XVI,  §  72. 

In  case  both  a  theorem  and  its  converse  are  true,  they 
may  be  stated  together  in  one  sentence  by  properly  using  the 
clause  "  if  and  only  if."  Thus,  Theorems  VII  and  VIII  may 
be  combined  into  one  as  follows :  "  When  two  lines  are  cut  by 
a  transversal,  the  alternate  interior  angles  are  equal  if  and 
only  if  the  lines  are  parallel." 

The  phrases  "and  conversely"  or  "and  vice  versa''  are  often 
used  for  the  same  purpose.  Thus  we  might  say,  "  If  a  body  is 
lighter  than  water,  it  will  float;  and  conversely.''  Or,  "If  a 
body  is  lighter  than  water,  it  will  float ;  and  vice  versa." 

EXERCISES 

1.  Make  a  true  statement  whose  converse  is  false. 

2.  Make  a  true  statement  whose  converse  is  true. 

3.  Write  out  the  illustration  you  used  in  Ex.  2  in  the  form 
of  a  single  sentence,  using  the  clause  "  if  and  only  if." 

4.  Rewrite  each  double  statement  on  this  page,  using  the 
phrase,  (a)  "if  and  only  if";  {b)  "and  conversely,"  (c)  "and 
vice  versa."     Which  are  true  and  which  are  false  ? 


54 


RECTILINEAR  FIGURES 


fl,§56 


Fig.  40 


56.  Theorem  IX.     If  two  lines  are  cut  by  a  transversal  and 
the  corresponding  angles  are  equal j 
the  lines  are  parallel. 

Given  AB  and  CD  cut  by  the 
transversal  ST  m.  such  a  way  that 
the  corresponding  angles  x  and  y 
are  equal. 

To  prove  that  AB  is  parallel 
to  CD. 

Proof.  Z.x  =  Zz,  10,  §31 

but  .  Z.x=  /.y,  Given 

therefore  Z.y  =  Z.z,  Ax.  9 

and  hence  also  AB  \\  CD.  §  51 

57.  Corollary  1.  If  two  lines  are  cut  by  a  transversal  and  the 
two  interior  angles  on  the  same  side  of  the  transversal  are  supple- 
mentary, the  lines  are  ixtrallel. 

58.  Corollary  2.  From  a  given  poi^it  only  one  perpendicular 
ca7i  be  drawn  to  a  given  line.  For,  if  there  were  two  perpen- 
diculars through  the  same  point,  they  would  be  parallel  (Why?) 
but  parallel  lines  cannot  meet  (§  48).     See  also  Ex.  2,  p.  48. 


EXERCISES 

1.  Two  parallel  pipes  for  hot  and  cold 
water  lie  flat  along  the  same  wall ;  at  the  end 
of  each  of  them  an  elbow  is  screwed  on  which 
turns  the  pipe  through  a  right  angle.  If  the 
pipes  connected  to  these  elbows  also  lie  flat 
against  the  same  wall,  will  they  be  parallel  ? 
Connect  your  answer  with  §§  56-58. 

2.  A  rectangle  (§  24)  has  all  its  angles  right  angles ;  show 
that  the  opposite  sides  are  parallel. 


I,  §60}  PARALLEL  LINES  55 

59.  Theorem  X.  {Converse  of  Theorem  IX.)  If  two  paral- 
lel lines  are  cut  by  a  transversal^  the  corresponding  angles  are 
equal. 

Given  the  two  parallel  lines  AB 
and  CD  cut  by  the  transversal 
ST^  making  corresponding  angles 
X  and  y. 

To  prove  that  /.x=/.y. 

Proof.  /.x=Z.z,     Why? 

also  /.y=/.z\    Why? 

therefore  Ax=Ay.    Why  ? 


Fio.  41 


60.  Corollary  1.  If  a  line  is  perpendicular  to  one  ofttuo  par- 
allelsy  it  is  perpendicidar  to  the  other  also, 

EXERCISES 

1.  State  and  prove  the  converse  of  Corollary  1,  Theorem 
IX. 

2.  The  crosspieces  (arms)  which  are  put  on  a  telephone 
pole  to  carry  the  wires  are  usually  all  perpendicular  to  the 
pole.     How  does  this  illustrate  Theorem  X  or  its  corollary  ? 

3.  Prove  that  the  diagonals  of  all  of  the  squares  on  a  sheet 
of  squared  paper  form  continuous  lines. 

4.  Prove  that  the  bisectors  of  any  pair  of  corresponding 
angles  formed  when  a  transversal  cuts  two  parallel  lines  are 
themselves  parallel.  Is  this  true  also  for  bisectors  of  alternate 
interior  angles  ? 

5.  Prove  that  the  bisectors  of  any  two  interior  angles 
formed  when  a  transversal  cuts  two  parallels  are  either  par- 
allel, or  else  perpendicular  to  each  other. 

6.  Prove  by  §§  56,  59  that  lines  perpendicidar  to  the  same 
line  {or  to  parallel  lines)  are  parallel,     (See  also  §  52.) 


56  RECTILINEAR  FIGURES  [I,  §61 

61.    Problem  1.     To  construct  a  line  parallel  to  a  giveii  line 
and  passing  through  a  given  point. 

T/ 

/ 

o/  K 


■ ? 

Fig.  42 

Given  the  line  AB  and  the  point  0. 

Required  to  construct  a  line  through  O  parallel  to  AB. 
(See  Ex.  2,  p.  50.) 

Construction.  Draw  any  line  OT  through  0,  cutting  AB  at 
some  point,  as  D. 

At  0  construct  Z  TOK  =  Z  ODB.  Prob.  5,  p.  6. 

Then  the  line  OK  (extended)  will  be  the  desired  line  through 
0  parallel  to  AB. 

Proof.  Since  Z  TOK=  Z  ODB,  the  lines  AB  and  OK  are 
parallel.  §  56 

Note.  The  fact  that  in  the  problem  of  §  61,  above,  we  have  not 
only  shown  how  to  draw  the  desired  line,  but  have  afterwards  proved 
that  our  method  is  correct,  illustrates  the  principle  that  every  construction 
problem  should,  in  its  solution,  contain  not  only  the  construction,  but  also 
the  proof  of  its  correctness.  This  will  be  done  hereafter  in  such  problems 
as  are  worked  out  in  the  text,  and  the  student  should  do  the  same  in  all 
construction  problems  that  occur  in  exercises. 

EXERCISES 

1.  Show  how  to  construct  a  line  parallel  to  a  given  line  and 
passing  through  a  given  point,  by  means  of  §  52. 

2.  Show  how  to  construct  a  line  that  makes  one  half  a  given 
angle  with  a  given  line  at  a  given  point. 

[Hint.    Mrst  bisect  the  given  angle.] 

3.  Show  how  to  construct  a  line  that  makes  an  angle  of  45° 
with  a  given  line  at  a  given  point. 


I,  §61] 


PARALLEL  LINES 


57 


4.  (a)  In  order  to  draw  a  parallel  to  a  line  I  through  a 
point  P,  a  draftsman  will  place  a  drawing  triangle,  or  other 
object  with  two  fixed  edges,  so  that  one  side  of  the  triangle 


coincides  with  Z,  and  the  other  side  passes  through  P.  He 
will  then  lay  a  ruler  against  the  side  of  the  triangle  that 
passes  through  P,  and  finally  slide  the  triangle  along  the  edge 
of  the  ruler,  until  one  corner  of  the  triangle  comes  to  P. 
Show  that  a  line  drawn  along  the  side  of  the  triangle,  originally 
in  coincidence  with  I,  will  be  the  parallel  to  I  through  P. 


(b)  Show  how  to  draw,  by  means  of  a  drawing  triangle,  a  per- 
pendicular to  a  given  line  AB  through  a  given  point.  (Fig.  b.) 
Notice  that  Z  i?  =  90°.     See  Ex.  3,  p.  41. 

5.  Draw  Fig.  38,  omitting  the  portion  dotted  in  the  figure. 
Through  some  point  C  on  CD  construct  a  line  parallel  to  ST. 
(The  quadrilateral  formed  is  called  a  parallelogram.) 

6.  A  parallelogram  (Ex.  5)  is  formed  when  one  pair  of 
parallel  lines  cuts  another  pair.  Show  that  the  sum  of  the 
two  interior  angles  that  have  one  side  in  common  is  180°. 


58  RECTILINEAR  FIGURES  [I,  §62 

PART   III.    ANGLES   AND  TRIANGLES 

62.    Theorem  XI.     The  sum  of  the  three  angles  of  any  tri' 
angle  is  equal  to  two  right  angles,  or  180^. 


Given  any  triangle  ABC. 

To  prove  that  /.  A+Z.B  +  ^  (7  =  2rt.  A. 

Proof.  Prolong  AB  to  D  and  through  B  draw  BE  \\  AC. 
Denote  Z  DBE  by  x,  and  Z  EBC  by  y. 

Then  Zx-^  Ay  +  Z  ABC  =  2  rt  A-,  Why? 

moreover  A  A  =  Ax,  §  59 

and  AC  =  Ay.  §54 

Therefore       Z  ^  -f-  Z  O  +  Z  ^.BO  =  2  rt.  A.  Ax.  9 

Note.  This  famous  theorem  is  of  great  practical  importance. 
It  was  known  by  Pythagoras  (about  500  e.g.).  This  figure  was 
used  by  the  Greek  philosopher  Aristotle  (384-322  b.c.)  and  by 
the  famous  Greek  geometrician  Euclid  (about  300  e.g.). 

63.    Corollary  1.     The  sum  of  the  two  acute  angles  of  any  right 
triangle  is  one  right  angle,  or  90°. 
f      64.    Corollary  2.     An  exterior  angle  of  any  triangle  is  equal  to 
the  sum  of  its  opposite  interior  angles.     That  is,  in  Fig.  43 

ACBD=ZA  +  ZC. 

65.  Corollary  3.  Each  angle  of  aii  equilateral  triangle  is 
equal  to  60°. 

66.  Corollary  4.  If  tico  angles  of  one  triangle  are  equal  re- 
spectively to  two  angles  of  another  triangle,  then  the  third  angles 
are  likewise  equal. 


I,  §661 


ANGLES  AND  TRIANGLES 


59 


EXERCISES 

1.  Two  angles  of  a  triangle  are  10°  30'  and  85°  15',  respec- 
tively.    What  is  the  size  of  the  third  angle  ? 

2.  If  the  rafters  of  the  roof  represented  in  Ex.  2,  p.  43, 
make  an  angle  of  35°  with  the  horizontal,  show  that  the  total 
angle  at  the  ridge  of  the  roof  is  110°. 

3.  A  crank  AB  is  operated  by  means  of  a  rod  DVB,  which 
slides  through  a  ring  at  C.     Show  that  g 
the  angle  ACB  is  always  half  the  angle                     ^^^^y 
XAB,  provided  AC  =  AB.                           -    ^^"^^^^  {  X 

4.  The   exterior   angles   at  A  and  C 

of  a  triangle  ABC  are  71°  and  140°,  respectively ;  how  many 
degrees  in  the  angle  B  ? 

5.  Eind  the  three  angles  of  an  isosceles  triangle  when  one 
of  the  angles  at  the  base  is  equal  to  one  half  the  angle  at  the 
vertex. 

6.  Draw  an  equilateral  triangle,  and  draw 
three  other  equilateral  triangles,  placing  one 
on  each  of  the  sides  of  the  first  one  as  a  base. 
Show  that  if  this  process  is  repeated  continu- 
ally, the  plane  is  divided  into  equilateral  tri- 
angles that  completely  fill  it.  This  fact  is  the  basis  for  many 
interesting  designs,  some  of  which  are  shown  below. 


l/f^plq 

i>Ki>< 

A/\/\A/\/yV\A/\7 

KAA/Ox^VAAAkX 

><>< 

N^i^lcTi 

Orchard  Plan  Tiled  Floor  Pattern  Linoleum  Pattern 

7.  Show  by  §  62  that  a  triangle  can  have  no  more  than  one 
obtuse  angle.  Can  it  have  more  than  one  right  angle  ?  See 
Exs.  2,  3,  p.  48. 


60  RECTILINEAR  FIGURES  [I,  §67 

67.  Theorem  XII.  Two   angles   whose  sides  are  re- 
spectively parallel 
are  either  equal  or 

supplementary,  ^^ ^<.'A__.      ^X        c 

There  are  two  cases,         i/    /      / X^ 

as  indicated  by  Figs.       ^  ^    ^  B 

44  a  and  44  6.  ^^«- ^«  Fig.  44& 

Case  1.  Given  A  B  and  B'  with  AB  \\  A'B'  and  BC  ||  5'0', 
as  in  Fig.  44  a. 

To  prove  ZB  =  ZB\ 

Proof.  Prolong  BC  and  B'A'  until  they  meet,  thus  form- 
ing the  angle  x. 

Then  ZB=  Zx,  Why  ? 

and  ZB'=Zx.  Why? 

Therefore  ZB  =  ZB'.  Why  ? 

Case  2.  Given  Zs^  and  B'  with  ^5  II  A'B'  and  50 1|  J5'C", 
as  in  Fig.  44  b. 

To  prove  that  A  B  and  B'  are  supplementary. 

[The  details  of  the  proof  for  this  case  are  left  to  the  student.] 

EXERCISES 

1.  When  in  Theorem  XII  will  the  two  angles  be  equal? 
when  supplementary? 

2.  Show  that  Theorem  XII  is  illustrated  by  the  angles  at 
the  intersection  of  any  two  straight  streets  of  uniform  width. 

3.  Show  that  Theorem  XII  is  illustrated  by  the  angles  at 
the  intersection  of  two  straight  railroads. 

4.  A  parallelogram  (Ex.  5,  p.  57)  is  a  figure  formed  when 
one  pair  of  parallel  lines  cuts  another  pair.  Show  by  means  of 
§  67  that  the  opposite  interior  angles  are  equal,  and  that  the 
adjacent  interior  angles  are  supplementary. 


I,  §681 


ANGLES  AND  TRIANGLES 


61 


68.  Theorem  XIII.  T%t)o  angles  ivhose  sides  are  re- 
sjjectively  jyerpendicular  to  each  other  are  either  equal 
or  supplementary. 

There  are  two  cases,  as  indicated  by  Figs.  45  a  and  45  b. 

Case  1.  Given  AB  and  B'  with  AB±B'a  and  BC±A'B\ 
as  in  Fig.  45  a. 

To  prove  that        ZB  =  ZB'. 

Proof.  Prolong  AB  and  B'C  until  they 
meet  at  some  point  such  as  T,  and  through 
T  draw  a  line  TT  II BC,  meeting  A'B'  pro- 
longed at  K. 

Then  Zx==Zy.         Why? 

Now  Zz  is  the  complement  of  Zy.     ' 

Moreover,  TKB'  is  a  right  angle  ; 
hence  Zz  is  the  complement  of  Z  B'. 

Therefore  Zy  =  Z  B',  8,  §  31 
or  ZB=ZB'.  Ax.  9 

Case  2.  Given  zi  S  and  B'  with 
^S±5'C'  and  BC±A'B',  as  in  Fig. 
456. 

To  prove  A  B  and  B'  supplementary. 

[The  details  of  the  proof  for  this  case  are 
left  to  the  student.] 


'B' 
Fig.  45  a 


Def.  §  18 
§60 

§a3 


EXERCISES 

1.  When  in  Theorem  XIII  will  the  two  angles  be  equal  ? 
when  supplementary  ? 

2.  An  object  lies  at  a  point  K  on  an  inclined 
plane  AB.  Show  that  the  angle  between  the 
vertical  line  through  K  and  a  perpendicular  to     '^ 

AB  at  K  is  equal  to  the  angle  CAB  which  the  inclined  plane 
makes  with  the  horizontal. 


62 


RECTILINEAR  FIGURES 


[I,  §69 


69.  Theorem  XIV.  Two  right  triangles  are  congruent  if 
the  hypotenuse  and  an  acute  angle  of  the  one  are  equal  respec- 
tively to  the  hypotenuse  and  an  acute  angle  of  the  other. 


Ftg.  46 

Given  the  rt.  A  ABC  and  A'B'C  with  the  hypotenuse  BC  = 

hypotenuse  B'C  srnd  Z  B=Z  B'. 

To  prove  A  ABC  ^  A  A'B'C. 

Proof.     We  have  Z:A  =  ZA'3indZB=ZB'.      Given 

Hence  ZC=ZC'.  ^66 

Moreover  BC=  B'C,  Given 

and  therefore  A  ABC  ^  A  A'B'C.  Why  ? 

Note.     Theorem  XIV  is  frequently  stated  in  the  following 

form :  A  right  triangle  is  determined  by  its  hypotenuse  and  one 

acute  angle. 

EXERCISES 

1.  Draw  an  acute  angle  and  then  construct  the  right  triangle 
containing  this  angle  and  having  a  hypotenuse  2  inches  long. 

2.  Show  that  if  two  right  triangles  have  one  acute  angle  of 
one  equal  to  one  acute  angle  of  the  other,  all  of  the  angles  of 
the' one  are  equal  to  the  corresponding  angles  of  the  other. 

3.  If  one  angle  of  a  right  triangle  is  45°,  show  that  the  tri- 
angle is  isosceles. 

4.  State  and  prove  the  converse  of  Ex.  3. 

5.  How  could  you  construct  (using  only  ruler  and  compasses) 
the  right  triangle  one  of  whose  acute  angles  is  60°  and  whose 
hypotenuse  is  a  given  length  AB  ?     (See  §  65.) 


I,  §71] 


ANGLES  AND  TRIANGLES 


63 


Fig.  47 


70.  Theorem  XV.  Two  right  triangles  are  congruent  if  the 
hypotenuse  and  a  side  of  the  one  are  equal  respectively  to  the 
hypotenuse  and  a  side  of  the  other. 

Given  the  rt.  A  ABC  and  A'B'C  with  hyp. 
BC=  hyp.  BfC  and  side  AB  =  side  A'B'. 

To  prove  that  rt.  A  ABC  ^  rt.  A  A'BC. 

Proof.  Place  A  A'B'C  in  the  position 
ABC"  so  that  A'B^  coincides  with  its  equal 
AB  and  C  falls  at  C'\  opposite  to  C. 

Then,  the  A  CAB  and  CAB  being  rt.  A 
(why?),  the  line  CAC"  will  be  a  straight 
line.  12,  §  31 

Now,  in  the  A  CC"B,  we  have 

BC=B'C"',  Given 

therefore  A  CC"B  is  isosceles ; 
hence  AACB  =  AAC"B, 

or  AACB=:AA'C'B'. 

Therefore  rt.  A  ABC  ^  rt.  A  A!BfC', 

71.  Corollary  '1.  If  two  oblique  lines  of  equal  length  are 
drawn  from  a  point  C  in  a  pei-pendicular  CD  to  a  line  AB  (Fig. 
32,  p.  42),  they  cut  off  equal  distances  from  the  foot  of  the  per- 
pendicular, and  conversely. 

EXERCISES 

1.  In  the  figure  a  mast  is  being  held  in  a  vertical  position 
by  means  of  a  number  of  ropes  (guy  ropes)  at- 
tached to  the  mast  at  the  same  distance  from  the 
ground.  Show  that  the  ropes  will  all  have  the 
same  length  if  they  are  anchored  at  equal  dis- 
tances from  the  foot  of  the  mast. 

2.  Construct  a  right  triangle  whose  hypotenuse 
is  4  inches  long  and  one  of  whose  sides  is  2  inches  long. 


§40 
Ax.  9 

§69 


64 


RECTinXEAR  FIGURES 


[I,  §72 


72.  Theorem  XVI.  IConverse  of  Theorem  IIL~\  If  two 
angles  of  a  triangle  are  equal,  the  sides  opposite  are  equals  and 
the  triangle  is  isosceles.  C 

Given  the  A  ABC  in  which  ZA=ZB. 
To  prove  that  AC  =  BC 

Proof.     Draw  CD  bisecting  Z  C. 
Then,  in  the  A  ACD  and  BCD  we  have 

Zx  =  ZyajidZ.A  =  ZB.  Why? 

Therefore  Zu  =  Zv:  §  66 


Fig.  48 


but 

CD=CD. 

Iden. 

Hence 

A  ACD  ^  A  BCD, 

Why? 

and  therefore 

AC=BC. 

Why? 

73.  Corollary  1.     -4n  equiangular  triangle  is  also  equUaterdl. 

74.  Corollary  2.  If  two  oblique  lines  are  draion  from  a  point 
C  in  a  perpendicular  CD  to  a  line  AB  (Fig.  48),  so  as  to  make 
equal  angles  with  AB,  they  are  equal. 

75.  Theorem  XVII.  If  two  angles  of  a  triangle  are  unequal, 
the  sides  opposite  them  are  unequal  and  the  greater  side  is  oppo- 


site  the  greater  angle. 

r 

Given  the  A  ABC  in  which  ZB> 

ZA. 

^^-^^    \ 

To  prove  that  AC  >  BC.                       ^  ^ 

-^B 

Fig.  49 

Proof.    Draw  BD  meeting  AC  at  D 

and  making  Zx=  ZA. 

Then,  in  the  A  ABD 

AD  =  BD, 

Why? 

and                                  BU  +  DC  >  CB. 

Post.  3 

Therefore                       AD  +  DC  >  CB ; 

Ax.  9 

that  is,                                AC>CB. 

76.   Corollary  1.     If  two  triangles  have  two  sides  of  the  owe 
eqiuil  to  two  sides  of  the  other,  hut  the  included  angle  of  the  first 


I,  §77]  ANGLES  AND  TRIANGLES  65 

greater  than  the  included  angle  of  the  second,  then  the  third  side 
of  the  first  is  greater  than  the  third  side  of  the  second. 

[Hist.  In  ^ABC  and  A'B'C,  \etAB=A'B',  BC=B'C',  And  Z  ABC 
>  Z  A'B<  a.  Suppose  AB  <  BC.  Place  A'B'  C  on  ABC  with  A' B>  on 
AB.  Join  CC.  Then  ABCC  is  isosceles,  and  ZBC'C  =  ZBCC'. 
Hence,  in  HAC'C,  ZAC'C>ZACC\  whence  (§  76)  AC>AC'.] 

Note.  Corollary  1  is  sometimes  stated  in  the  following 
brief  form  wherein  it  finds  numerous  illustrations  in  physics 
and  mechanics.  "  The  growth  of  an  angle  in  a  triangle  means 
the  growth  of  the  side  opposite  it.''  It  is  to  be  understood, 
of  course,  in  this  statement,  that  as  the  angle  is  allowed,  to 
grow,  the  lengths  of  its  including  sides  remain  fixed. 

77.  Corollary  2.  If  two  oblique  lines  are  drawn  from  a  point 
C  in  a  perpendicular  CD  to  a  line  AB  (Fig.  48),  and  if  the 
base  angles  at  A  and  B  are  unequal,  the  oblique  line  opposite  the 
greater  base  angle  is  the  greater;  in  particular,  the  perpendicular 
CD  is  itself  the  shortest  line  from  C  to  any  point  of  AB. 


EXERCISES 

1.  If,  on  account  of  some  obstruction,  one  of  the  guy  ropes 
mentioned  in  Ex.  1,  p.  63,  must  be  anchored  nearer  the  foot  of 
the  mast  than  the  others,  show  why  that  rope  will  be  the 
shortest. 

2.  Is   the   string  attached   to    a  kite   usually         \  ^^^ 
equal  in  length  to  the  height  of  the  kite  above 
the  ground  ?     Connect  your  answer  with  §  75. 

3.  A  simple  form  of  crane  consists  of  a  beam 
AB  hinged  at  ^  to  a  vertical  mast  AC  and  con- 
trolled by  a  wire  rope  attached  at  B  and  run-      C""*"® 
ning  over  a  pulley  at  C.     When  the  rope  is  let  out,  the  beam 
AB  descends.     Connect  this  fact  with  §  76. 


66  RECTILINEAR   FIGURES  [I,  §78 

78.  Theorem  XVIII.  If  two  sides  of  a  triangle  are  unequal, 
the  angles  opposite  them  are  unequal  and  the  greater  angle  is  op- 
posite the  greater  side. 

C 


Fig.  50 

Given  A  ABC  in  which  BC  >  AC. 

To  prove  that  Z  CAB  >  Z.B. 

Proof.     On  CB  take  CB  =  AC',  draw  AD, 

Then,  in  the  triangle  ADC  we  have 

ZCAD=ZCDA.  Why? 

But  ZCDA>ZB',  §47 

therefore  Z  CAD  >  Z  B.  Why  ? 

Moreover  ZCAB>  ZCAD-,  Ax.  10 

therefore  Z  CAB  >  Z  B.  Why  ? 

79.  Corollary  1.  If  two  triangles  have  tivo  sides  of  the  one 
equal  to  two  sides  of  the  other,  but  the  third  side  of  the  first  greater 
than  the  third  side  of  the  second,  then  the  included  angle  of  the 
first  is  greater  than  the  included  angle  of  the  second. 

[Hint.  Prove  by  reduction  to  absurdity.  Sliow  first  that  equality  of 
the  included  angles  leads  to  a  violation  of  §  35.  Show  that  if  the  included 
angle  of  the  second  triangle  is  the  greater,  §  76  is  violated.] 

Note.  Corollary  1  is  sometimes  stated  in  the  following- 
brief  form  wherein  it  finds  numerous  illustrations  in  physics 
and  mechanics :  "  The  growth  of  a  side  of  a  triangle  means  the 
growth  of  the  angle  opposite." 

80.  Corollary  2.  If  from  a  point  C  in  a  perpendicular  CD  to 
a  line  AB  (Fig.  48)  unequal  oblique  lines  are  drawn  to  the  base 
AB,  the  longer  of  the  oblique  lines  is  opposite  the  larger  of  the  two 
base  angles. 


I,  §80] 


ANGLES  AND  TRIANGLES 


67 


MISCELLANEOUS    EXERCISES 

1.  Prove  that  the  hypotenuse  of  a  right  triangle  is  its  longest 
side. 

2.  By  use  of  Theorem  XVIII,  prove  that  the  perpendicular  is 
the  shortest  line  that  can  be  drawn  from  a  point  to  a  straight 
line.     (Compare  §  77.) 

3.  If  the  crank  AB  mentioned  in  Ex.  3,  p.  59,  is  so  arranged 
that  AC  >  AB  (see  the  figure,  p.  59),  show  that  the  angle  ACB 
will  always  remain  less  than  half  the  angle  XAB  during  the 
rotation. 

4.  If  the  pans  of  a  balance  of  the  ordi- 
nary form  shown  in  the  figure  are  not  pre- 
cisely on  the  same  level,  show  that  each  pan 
is  nearer  to  the  middle  post  than  when  the 
balance  is  level.  Show  also  that  the  pans 
are  always  at  equal  distances  from  the 
middle  post. 


Balance  Scales 


King.Post  Truss 


5.  A  simple  piece  of  bridge  work  consists  of  a  frame  like 
that  shown  in  the  adjoining  figure,  AM  and  MB  being  stiff 
pieces  of  steel  merely  hinged  together 
at  M,  but  the  hinge  resting  on  a  plate 
P,  which  at  some  lower  point  C  is 
connected  to  A  and  B  by  strong  flexi- 
ble wires.  Show  that  a  heavy  weight 
may  safely  be  put  at  Jf,  even  though 
the  bridge  is  supported  only  by  buttresses  at  A  and  B.  (An 
arrangement  of  this  kind  is  called  an  inverted  King-Post 
truss.) 

6.  Determine  the  angles  of  a  triangle  when  they  are  in  the 
ratio  3:4:5. 

7.  If  the  exterior  angle  at  A  in  a  triangle  ABC  is  115°, 
and  C  is  three  times  B,  find  B  and  C 


68 


RECTILINEAR  FIGURES 


[I,  §81 


PART  IV.     QUADRILATERALS 

81.  Definitions.  A  plane  figure  bounded  by  four  straight 
lines  is  called  a  quadrilateral.  It  is  desirable  to  distinguish  be' 
tween  several  kinds  of  quadrilaterals  as  follows : 

If  two  sides  of  a  quadrilateral  are  parallel,  it  is  called  a 
trapezoid. 

If  each  pair  of  opposite  sides  of  a  quadrilateral  are  parallel, 
it  is  called  a  parallelogram. 


B     H 


Quadrilateral 


Trapezoid 
Fig.  51 


Parallelogram 


A  parallelogram  all  of  whose  angles  are  right  angles  is  called 
a  rectangle. 

A  quadrilateral  all  of  whose  sides  are  equal  is  called  a 
rhombus. 

A  rectangle  all  of  whose  sides  are  equal  is  called  a  square. 


Rectangle 


Rhombu 
Fig.  52 


Square 


The  side  upon  which  a  quadrilateral  appears  to  rest  is  called 
its  base.  Trapezoids  and  parallelograms,  however,  are  consid- 
ered as  having  two  bases,  one  being  the  side  upon  which  the 
figure  appears  to  rest,  and  the  other  being  the  parallel  side 
opposite  it.     Thus,  in  Fig.  53,  AB  and  CD  are  bases. 


I,  §81] 


QUADRILATERALS 


69 


The  perpendicular  distance  between  the  bases  (prolonged  if 
necessary)  of  a  trapezoid  or  parallelogram  is  called  its  altitude, 
as  the  line  h  in  the  figures  below. 


B       A 


D 

\  1 '-" 

f 

X     / 

^     \ 

\ 

1  \ 

/X 

N  \ 

\ 

I 

_J 

Fig.  53 


A  line  joining  opposite  corners  (vertices)  of  a  quadrilateral 
is  called  a  diagonal,  as  the  line  AG  in  Fig.  53. 

In  the  figures  above,  the  angles  ABC  and  BCD  are  said  to 
ije  adjacent  to  the  side  BC.  A  pair  of  angles  such  as  A  and 
C  are  said  to  be  oj^osite  angles  of  the  quadrilateral. 


EXERCISES 

1.  Prove  that  the  angles  adjacent  to  any  side  of  a  parallelo- 
gram are  supplementary. 

2.  Prove  that  opposite  angles  of  a  parallelogram  are  equal. 

3.  If  two  adjacent  angles  of  a  parallelogram  are  in  the  ratio 
17 : 1,  how  large  is  each  angle  of  the  parallelogram  ? 

4.  Construct  the  rhombus  each  of  whose  sides  equals  2  inches 
and  one  of  whose  angles  is  30". 

5.  Show  that  two  equilateral  triangles  that  have  a  common 


side,  together  form   a  rhombus.     This  is  popularly  called  a 
"  diamond,"  and  is  the  basis  of  many  designs. 


70 


RECTILINEAR  FIGURES 


[I,  §82 


82.   Theorem   XIX.      Either  diagonal  of  a  parallelogram 
divides  it  into  two  congruent  triangles. 

Given  the  parallelogram 
ABGD  in  which  the  diagonal 
AC  has  been  drawn. 

To  prove  that 

A  ABC  ^  A  ADC. 

Proof.  IniliQ  A  ABC,  ADC 
we  have  AC  =  AC, 

and  /.z  =  Z.w\ 

therefore  A  ABC  ^  ADC. 


Fig.  54 


83.   Corollary  1. 

side  opposite  it. 


§54. 

Why? 

Why? 

Any  side  of  a  parallelogram  is  equal  to  the 


84.  Corollary  2.  The  segments  of  par- 
allel lines  included  between  parallel  lines 
are  equal. 

[Thus,  in  Fig.  55,  AB  and  CD  are  one  pair 
of  parallels,  while  A'B'  and  CD'  are  another 
pair ;  they  form  the  quadrilateral  whose  sides 
are  represented  by  a,  &,  c,  and  d.  Then,  the 
corollary  states  that  a  =  c  and  h  =  c?.] 


EXERCISES 

1.  How  does  the  ruled  paper  used  in  drawings  in  the  Intro- 
duction (see  §  25)  provide  an  illustration  of  Corollary  2  ? 

2.  In  the  parallelograms  that  occur  in  the  framework  of 
bridges,  a  crosspiece  is  usually  inserted  along  at  least  one  of  the 
diagonals.    Why  will  this  make  the  whole  parallelogram  stiff  ? 

3.  Cut  a  piece  of  paper  in  the  form  of  a  parallelogram  and 
then  cut  it  in  two  along  one  diagonal.  Will  the  two  triangles 
thus  formed  fit  exactly  upon  each  other  ?     Why  ? 


I.  §86] 


QUADRILATERALS 


71 


85.   Theorem   XX.     If  a  quadrilateral  has  both  pairs  of 
opposite  sides  equal,  it  is  a  parallelogram. 


Why? 
Why? 
Why  ? 
Why? 
§81 


Fia.  56 

Given  the  quadrilateral  ABCD  in  which  AB  =  DC  and  BC 
=  AD. 
To  prove  that  ABCD  is  a  EJ. 

Proof.     In  the  A  ABC,  ADC  we  have 

AC=AC,  AB  =  DC,  and  BC  =  AD. 

Therefore  A  ABC  ^AADC-, 

hence  Z  x  =  Z.  y,  and  Z  z  =  Zw. 

It  follows  that  AB  II  DC,  and  BCWAD-, 
hence  ABCD  is  a  O. 


86.  Theorem  XXI.     If  a 

quadrilateral  has  one  pair  of 
sides  equal  and  parallel,  it  is  a 
parallelogram. 

Given  the  quadrilateral 
ABCD  in  which  ^B  is  equal 
and  parallel  to  DC. 

To  prove  that  ABCD  is  a  O. 

Proof.     Draw  the  diagonal  AC. 
ADC,  we  have 

AC=AC,AB  =  DC,  Zx 
hence  A  ABC  ^  A  ADC 

Therefore       Z  z  =  Zio,  and  hence  AD 

But  since  AB  II  DC,  ABCD  is  a  O. 


Fio.  57 


Then  in  the  A  ABC  and 


^y, 

Why? 

Why? 

BC. 

Why? 

Why? 

72 


RECTILINEAR  FIGURES 


[I,  §87 


87.  Theorem  XXII.  The  diagonals  of  a  parallelogram  bi- 
sect each  other. 

Given  the  O  ABCD,  and  let  its  /^\m 

diagonals  intersect  at  M. 

To  prove  that  AM=  MG  and  that     . 

DM=MB.  t.      .«     ^ 

Fig.  58 

[Hint.    Prove  that  A  AMB  ^  A  DMC. 

Since  the  proof  is  easily  carried  out  it  is  left  to  the  student.] 

88.  Theorem  XXIII.  Two  parallelograms  are  congruent  if 
two  sides  and  the  included  angle  of  the  one  are  equal  respectively 
to  two  sides  and  the  included  angle  of  the  other. 


■^B 


Fig.  59 


Given   the  n  ABCD  and   A'B'C'D'  in   which  AB  =  A'B\ 
AD  =  A'D',  and  Z  DAB  =  Z  D'A'B'. 
To  prove  O  ABCD  ^  O  A'B'C'D'. 

[Hint.    Draw  the  diagonals  DB,  D'B',  and  prove  A  ADB  ^  A  A'D'B'; 
also  prove  that  A  DCB  ^  A  D'C'B'.    Then  apply  §  33.] 


EXERCISES 

1.  Prove  that  the  diagonals  of  a  rectangle 
are  equal. 

2.  Prove  that  the  diagonals  of  a  rhombus  are 
perpendicular  to  each  other. 

3.  Show  that  if  each  of  the  diagonals  of  all 
of  the  squares  on  a  piece  of  squared  paper  are 
drawn,  two  new  sets  of  continuous  straight 
lines  at  right  angles  to  each  other  are  formed. 
This  is  the  basis  of  many  designs. 


I,  §90]  QUADRILATERALS  73 

89.  Theorem  XXIV.  The  line  joining  the  middle  points  of 
the  two  sides  of  a  triangle  is  parallel  to  the  base  and  equal  to 
half  the  base. 


Given  the  triangle  ABC  and  the  line  DE  joining  the  mid- 
points of  the  sides  AC  and  BC. 

To  prove  that  DE  II  AB,  and  that  DE  =  AB/2. 

Proof.     Draw  BB'  II  AC  meeting  DE  prolonged  at  F. 

Then,  in  the  A  DEC  and  EBF,  we  have 

CE  =  EB,Zr  =  Zs,Zx  =  Zy.  Why? 

Therefore,  A  DEC ^  A  EBF-,  hence,  DC  =  BF.  Why  ? 

But  DC  =  AD',  hence,  BF  =  AD,  and  ABFD  is  a  O.    Why  ? 

It  follows  that  DE  II  AB.  §  81 

The  fact  that  DE  =  AB/2  may  now  be  established  as  follows  : 

Since,  as  just  shown,  ABFD  is  a  O,  we  have 

DF=AB,  Why? 

But  DF=DE  +  EF. 

Moreover  DE  =  EF,  since  A  DEC  ^  A  ^Bi^. 

It  follows  that  DF=2  DE,  ov  AB  =  2  DE, 
that  is  DE  =  AB/2.  Ax.  4 

90.  Corollary  1.  (Converse  of  %  89.)  The  line  draivn  through 
the  middle  jyoint  of  one  side  of  a  triangle  parallel  to  the  base 
bisects  the  other  side. 

[Hint.  Draw  the  parallel;  and  draw  the  line  connecting  the  middle 
points  of  the  two  sides.  If  these  do  not  coincide,  show  by  §  89  that  §  49  is 
violated.    For  another  proof,  see  Ex.  1,  p.  75.] 


74  RECTILINEAR  FIGURES  [I,  §91 

91.  Theorem  XXV.  If  three  parallel  lines  cut  off  two  equal 
portions  of  one  transversal,  they  cut  off  two  equal  portions  of 
any  other  transversal, 

T        V 
E       R'/       n\     F 
C 
A 


Given  the  three  parallel  lines  AB,  CD,  and  EF',  let  >ST  be  a 
transversal  of  which  the  two  portions  PQ  and  QR  cut  off  by 
the  three  parallels,  are  equal. 

To  prove  that  the  three  parallels  cut  off  equal  portions  LM 
and  MN  on  any  other  transversal  UV. 

Proof.  ■  Through  It  draw  a  line  RK  II  UV  cutting  CD  at  J. 

Then  KJ:=  LM,  and  JR  =  MN-,  §  84 

but  KJ=JR\  §90 

hence  LM—  MN.  Ax.  9 

92.  Corollary  1.  If  a  series  of  parallel  lines 
cut  off  equal  portions  of  one  transversal,  they  cut 
off  equal  portions  of  any  other  transversal. 

[Hint.     Show  that  the  portions  cut  off  on  any  trans- 
versal are  equal,  taking  two  at  a  time,   by  Theorem       s  U 
XXV.]                                                                                                     Fig.  62 

93.  Corollary  2.     If  three  parallel  lines  cut  off  two  portions 
of  one  transversal,  one  of  which  is  double  the  other,  j  V 
they  cut  off  two  portions  of  any  other  transversal,         -6^ — ly 
one  of  which  is  double  the  other.                                      "         "" 


7 

V 
K 

V 

J 

y 

I 

V 

P 

/ 

\    ■ 

[Hint.  First  draw  a  fourth  parallel  through  the  mid- 
dle point  of  the  larger  portion  of  the  first  transversal,  and 
then  use  Cor.  1,  §  92.] 


I,§W1 


QUADRILATERALS 


75 


94.  Corollary  3.  If  three  parallel  lines  cut  off  two  portions  oj 
one  trayisversal,  one  of  which  is  n  times  the  other,  they  cut  off  two 
portions  of  any  other  transversal,  one  of  which  is  n  iimes  the  other. 


EXERCISES 

1.  Prove  that  the  line  DE  drawn  through 
the  middle  point  of  one  side  of  a  triangle 
ABC  parallel  to  the  base  bisects  the  other 
side  by  drawing  BF  ||  AC  and  showing  that 
A  DCE  ^  A  FBE.     Compare  §  90. 

2.  Prove  that  the  lines  joining  the  middle 
points  of  the  three  sides  of  a  triangle  divide  it 
into  four  congruent  triangles.  /  2 

[Hint.     Prove  A1^A2^A3^A4.] 
.C 

3.  Prove  that  perpendiculars  drawn  from 
the  middle  points  of  two  sides  of  a  triangle 
to  the  third  side  are  equal. 


4.  Prove  that  the  lines  joining  the 
middle  points  of  the  sides  of  any  quadri- 
lateral form  a  parallelogram. 

[Hint.  Draw  the  diagonals  of  the  original 
quadrilateral,  and  use  §  89.] 

5.  A  long  board  7^  in.  wide  is  to  be  cut  into  4  equal  parallel 
strips.  Show  that  it  can  be  marked  ready  for  sawing  in  the 
following  manner : 

Lay  the  corner  (heel)  of  a  carpen- 
ter's square  on  one  edge  of  the  board 
(see  figure)   and  turn  until   the  12  v 

mark  is  on  the  other  edge.  With  an  awl  make  dents  at  3,  6, 
and  9.  Move  the  square  and  repeat  the  operation.  Then  draw 
parallels  through  the  dents  thus  made.     Verify  this  process. 


76  RECTILINEAR  FIGURES  [I,  §95 

PART  y.     POLYGONS 

95.  Definitions.  A  plane  figure  bounded  by  any  number  of 
straight  lines  is  called  a  polygon.  The  bounding  lines  are 
called  sides  ^  an  angle  between  two  adjacent  sides,  as  the  angle 
ABG  in  the  figure,  is  called  an  interior  angle  of  the  polygon, 
while  an  angle  between  any  one  side  and  the  adjacent  side  pro- 
longed, as  the  angle  CBK  in  the  figure,  is  called  an  exterior 
angle  of  the  polygon.  In  what  follows,  we  assume  that  all 
polygons  mentioned  are  convex,  i.e.  that  each  interior  angle  is 
less  than  180°. 


E 


A  line  joining  any  two  non-consecutive  vertices  is  a  diagonal, 
as  AC  in  the  figure. 

96.  Kinds  of  Polygons. 

A  triangle  is  a  polygon  of  three  sides. 

A  quadrilateral  is  a  polygon  of  four  sides. 

A  pentagon  is  a  polygon  of  five  sides. 

A  hexagon  is  a  polygon  of  six  sides. 

An  octagon  is  a  polygon  of  eight  sides. 

A  decagon  is  a  polygon  of  ten  sides. 

An  equilateral  polygon  is  one  all  of  whose  sides  are  equal. 

An  equiangular  polygon  is  one  all  of  whose  interior  angles 
are  equal. 

A  regular  polygon  is  one  which  is  both  equilateral  and  equi- 
angular. 


I,  §97]  POLYGONS  77 

97.  Theorem  XXVI.  The  sum  of  the  interior  angles  of  a 
polygon  is  two  right  angles  taken  as  many  times  as  the  figure  has 
sides,  less  two.  ^ 

Given  the  polygon  ABCD  •  •  •  having 
n  sides.     [In  Fig.  65,  n  =  6.] 

To  prove  that  the  sum  of  its  inte- 
rior angles  =  (n  —  2)  2  rt.  A. 

Proof.  Draw  the  diagonals,  AC, 
AD,  ••',  dividing  the  polygon  into 
(n-2)A.  Fig.  65 

The  sum  of  the  angles  of  the  polygon  is  equal  to  the  sum  of 
the  angles  of  these  triangles.  Ax.  11 

But,  the  sum  of  the  angles  of  any  triangle  =  2  rt.  zi,  §  62 ; 
hence  the  sum  of  the  angles  of  ABCD  —  is  (n  —  2)  2rt.  A. 

EXERCISES 

1.  "What  is  the  sum  of  the  interior  angles  of  a  pentagon  ?  a 
decagon  ?     an  octagon  ? 

2.  How  many  degrees  in  one  angle  of  a  regular  pentagon  ? 
Answer  the  same  question  for  a  regular  hexagon ;  regular  deca- 
gon ;  regular  octagon. 

3.  How  many  sides  has  the  polygon  each  of  whose  exterior 
angles  equals  30°  ? 

4.  Show  that  a  regular  hexagon  can  be  made  by  placing  six 
equilateral  triangles  with  one  vertex  of  each  at  the  same  point. 
(See  Ex.  6,  p.  59.) 

5.  Show  that  if  a  regular  hexagon  is 
Irawn  on  each  side  of  a  given  regular 
hexagon,  the  space  in  the  plane  around 
the  given  hexagon  is  just  tilled. 

If  the  process  is  continued,  show  that 
the  entire  plane  is  divided  into  regular 
hexagons,  in  the  manner  of  a  honeycomb. 


78  RECTILINEAR  FIGURES  [I,  §98 

98.  Theorem  XXVII.  Tlie  sum  of  the  exterior  angles  of  a 
polygon  formed  by  producing  the  sides  in  succession  is  equal  to 
four  right  angles. 

Given  the  polygon  ABCD  —.  ""dTo c\ 

To  prove  that  the  sum  of  its  exterior        /  \ 

angles  =  4  rt.  A.  /'eV  ^V^'^ 

Proof.     Denote  the  interior  A  by  A,  B,         \L__A^__ 
C,  D,  •••  and  the  corresponding  exterior  \ 

angles  by  a,  h,  c,  d,  •••.  "    Fig.  66 

Then,  A  A  +  A  a  =^2  rt.  A, 

AB-\-Ab  =  2vtA. 

In  like  manner  the  sum  of  each  pair  of  angles  at  a  vertex  = 
2  rt.  A. 

Therefore,  the  sum  of  both  interior  and  exterior  angles  about 
the  whole  polygon  will  be  2  rt.  A  taken  as  many  times  as  the 
polygon  has  sides ;  that  is,  it  will  be  «  •  2  rt.  A. 

But  the  sum  of  the  interior  angles  alone  is  (n  —  2)  •  2  rt.  A. 

§  97 
Therefore,  the  sum  of  the  exterior  angles  alone  is 

n  •  2  rt.  Zs  -  (n  -  2)  .  2  rt.  Zs  =  2  .  2  rt.  Zs  =  4  rt.  Z. 

EXERCISES 

1.  What  is  the  sum  of  the  interior  angles  of  a  square? 
What  is  the  sum  of  the  exterior  angles  ? 

2.  What  is  the  sum  of  the  interior  angles  of  any  quadri- 
lateral ?  What  is  the  sum  of  the  exterior  angles  ?  Compare 
with  Ex.  1. 

3.  What  is  the  sum  of  the  exterior  angles  of  a  pentagon  ? 
of  a  hexagon  ? 

4.  How  large  is  each  of  the  exterior  angles  of  a  regular 
pentagon  ?  hexagon  ?  decagon  ? 

5.  How  many  sides  has  the  polygon  the  sum  of  whose 
interior  angles  equals  the  sum  of  its  exterior  angles  ? 


I,  §99]  LOCUS  OF  A  POINT  79 


PART  VT.     THE  LOCUS   OF  A  POINT 

99.   Locus  of  a  point.     In  §  23  (Introduction)  it  is  stated 
that  a  circle  is  a  curve  every  point  of  which  is  equally  distant 
from  a  point  within  (center).     This  definition  may  be  stated  in 
the  following  language:  A  circle  is  the  locus  (position)  of  all 
points  equidistant  from  a  given  point.    Using  the  same  lan- 
guage, it  may  be  said  that  the  locus  of  all 
points  equidistant  from  two  parallel  lines     ^' 
is  the  line  lying  midway  between  them,  as      ^ 
the  line  EF  in  the  figure.  ^  ~ 

Similarly,  the  locus  of  all  points  common 
to  two  intersecting  lines  is  simply  one  point;  namely,  their 
point  of  intersection. 

Note.  It  is  important  to  observe  that  in  each  of  the  pre- 
ceding illustrations,  the  locus  not  only  (1)  contains  all  points  that 
satisfy  a  certain  given  condition,  hut  it  is  also  true  that  (2)  there 
are  no  points  on  the  locus  that  do  not  satisfy  this  condition. 
Thus,  in  the  figure  above  we  can  make  the  following  two 
statements  about  EF:  (1)  EF  contains  all  points  equidistant 
from  AB  and  CD ;  (2)  there  are  no  points  on  EF  that  are  not 
equidistant  from  AB  and  CD. 

Every  true  locus  ]?ossesses  the  properties  (1)  and  (2) ;  hence 
in  all  locus  problems  two  things  are  to  be  proved.  This 
will  be  illustrated  presently. 


EXERCISES 

1.  What  is  the  locus  of  all  points  2  inches  distant  from  a 
given  straight  line  ? 

[Hint.    Note  that  there  are  such  points  on  either  side  of  the  given  line.] 

2.  What  is  the  locus  of  points  1  in.  from  a  fixed  point  ? 

3.  What  is  the  locus  of  all  points  4  inches  distant  from  each 
of  two  given  points  which  are  6  inches  apart  ? 


80 


RECTILINEAR  FIGURES 


[I,  §  100 


100.  Theorem  XXVIII.     The  locus  of  all  points  equidistant 
from  the  extremities  of  a  line  is  the  perpendicular  bisector  of 

that  line. 

E 


Given  the  line  whose  extremities  are  A  and  B.     Given  also 
the  line  EC  drawn  _L  AB  at  its  middle  point  C. 

To  prove  that  EC  is  the  locus  of  all  points  equidistant  from 
A  and  B ;  that  is  (see  §  99),  to  prove  that 

(1)  any  point  D  which  is  equidistant  from  A  and  B  lies  on  EC, 

(2)  there  is  no  point  on  EC  not  equidistant  from  A  and  B. 

Proof.     From  D  draw  DA  and  DB. 

Then  A  ADC  ^  A  BDC.  §  45 

Therefore  Z  ACD  =  Z  BCD,  Why  ? 

so  that  DC±AB.  Why? 

Hence  D  must  lie  on  EC,  which   is  the  property  (1)  to  be 
proved.  T,  §  31 

Again,  let  D'  represent   any   point   on   the   perpendicular 
bisector  CE. 

Then  AAD'C^A  BCD'.  §  35 

Hence  D'A  =  D'B, 

which  is  the  property  (2)  to  be  proved. 


EXERCISE 

1.   What  is  the  locus  of  the  vertices  of  all  isosceles  triangles 
constructed  on  a  given  base  ? 


I,  §  101] 


LOCUS  OF  A  POINT 


81 


101.   Theorem  XXIX.     The  bisector  of  an  angle  is  the  locus 
of  all  points  equidistant  from  its  sides. 


Given  the  angle  ABC  and  its  bisector  BE. 
To  prove  that  BE  is  the  locus  of  all  points  equidistant  from 
AB  and  BC\  that  is,  to  prove  that 

(1)  any  point  D  equidistant  from  AB  and  BC  lies  on  BE, 

(2)  any  point  i>'  on  BE  is  equidistant  from  AB  and  BC. 
Proof.     For  the  proof  of  (1)  show  that  A  BDF^A  BDG. 

For  the  proof  of  (2)  show  that  A  BDF'  ^  A  BD'G\ 

[The  details  of  the  proof  are  left  as  an  exercise.] 

EXERCISES 

1.  By  means  of  Theorem  XXVIII  prove  the  correctness  of 
the  construction  given  in  §  5,  p.  4.  Similarly,  prove  the  cor- 
rectness of  the  constructions  indicated  in  §§  6,  8. 

2.  By  means  of  Theorem  XXIX  prove  the  correctness  of  the 
construction  given  in  §  9,  p.  8. 

3.  What  is  the  locus  of  a  point  that  is  equidistant  from 
three  given  points  ?    Show  how  to  construct  the  locus. 

4.  A  carpenter  bisects  an  Z  ^  as  follows : 
Lay  off  AB  =  AC.  Place  a  steel  square  so 
that  BD=  CD  as  shown  in  the  figure.  Mark 
D  and  then  draw  the  line  AD.  Show  that  AD 
bisects  Z  A.  Would  this  method  be  correct 
if  the  square  were  not  a  right  angle  at  D  ? 


82 


RECTILINEAR  FIGURES 


[I,  §  102 


102.  Supplementary  Theorems  on  Altitudes,  Medians,  etc. 

Theorem  XXX.       The  perpendiculars  erected  at  the  middle 
points  of  the  sides  of  a  triangle  meet  in  a  point. 

C 


Outline  of  proof.  Let  0  be  the  point  where  the  perpendicular  bisectors 
EI,  FH,  of  the  sides  AB,  BO  meet.  Join  O  to  the  middle  point  D  of 
AC.     Prove  that  OD  is  then  the  perpendicular  bisector  of  AC. 

Theorem  XXXI.  The  bisectors  of  the  angles  of  a  triangle 
meet  in  a  point. 

C 

D 


Fig.  71 

Outline  of  proof.  Draw  the  bisectors  of  A  A  and  B  and  suppose  that 
they  meet  at  0.  Join  O  to  the  third  vertex  0  and  prove  that 
rt.  A  OCD  ^  rt.  A  OCE,  thus  making  Z  ECO  =:ZDCO. 

Theorem  XXXII.     TJie  altitudes  of  a  triangle  meet  in  a  poijit. 

Outline  of  proof .    GivenAvl^C.    Through 

its  vertices  draw    lines  parallel  respectively    qI 

to  the   opposite    sides,    forming    A  A'B'  C        ^\ 
Then  A    is    the    mid-point    of  B'C,    since  v 

BO  AC  and  BOB' A  are  parallelograms. 
Similarly,  O  and  B  are  mid-points  of  A'B' 
and  A'C.  Then,  AD,  BE,  and  OF  are  per- 
pendicular to  the  sides  of  A'B'C  at  their 
mid-points  and  therefore  meet  in  a  point 
(Theorem  XXX). 


FiQ.  72 


I,  §  102]  MISCELLANEOUS  EXERCISES  83 

Theorem  XXXIII.  The  medians  of  a  triangle  meet  in  a  point 
which  is  tico  thii'ds  of  the  distance  from  any  vertex  to  the  middle 
point  of  the  opposite  side. 

Outline  of  proof.  Draw  the  medians  CF 
and  BE  and  suppose  they  intersect  at  0. 
Draw  AO  and  extend  it  to  cut  BC  at  D. 
Now  draw  BG  parallel  to  FC,  meeting  AO 
(prolonged)  at  G.  Johi  G  and  C.  Then  ^*^^^. 
in  the  triangle  ABG  we  have  AF  =  FB.  """-A-''" 

Therefore,    AO  =  OG.      (§  90,    p.    73.)  ^ 

Next  prove    OE  II  GO.     Then  BOCG  is  a  ^^^'  '^^ 

parallelogram.     Then  BD  =  DC  (§  87)  which  proves  that  the  medians 
meet  in  a  point. 

Moreover,  A0  =  OG,  while  0G  =  2  0D  (§  87).  Whence,  A0z=2  OD. 
But  AD=A0+  OD=S  OD  so  that  AO/AD  =  2/3 ;  that  is,  AO  =  i  AD. 


MISCELLANEOUS   EXERCISES   ON   CHAPTER  I 

1.  Given  two  angles  of  a  triangle.      Construct  the  third 
angle. 

2.  Prove  that  the  bisectors  of  two  supplementary  adjacent 
angles  are  perpendicular  to  each  other. 

3.  If  a  weight  is  hung  from  a  small  ring  that  slips  freely 
on  a  cord,  and  if  the  cord  is  tied  fast  to  two  sup- 
ports at  equal  heights,  the  ring  will  come  to 
rest  at  the  middle  of  the  cord.  Prove  that  the 
string  by  which  the  weight  is  attached  then 
bisects  the  angle  between  the  two  portions  of  the  cord. 

4.  A  simple  form  of  carpenter's  level  consists  of  three 
pieces  of  board  nailed  together  in  the  shape 
of  a  capital  letter  A.  A  plumb  bob  is  hung 
on  a  hook  screwed  at  B.  Show  that  any 
object  upon  which  the  feet  A,  C  are  set  will 
be  level  in  case  the  plumb  line  passes  through 
the  middle  point  Q  of  DE. 


84 


RECTILINEAR  FIGURES 


[I,  §  102 


5.  An  angular  joint  for  water  pipes  is  to  be  constructed  for 
two  pipes  that  meet  each  other  at  an  angle    q    \  h  / 
of  40°.     If  the  seam  FII  is  to  make  equal 
angles  with  the  lines  of  centers ;  that  is,  if 
angle  50i^  =  an gleJOF,  show  that  each  of 
these  angles  must  be  taken  equal  to  70°. 

6.  Prove  that  if  two  angles  of  a  quadrilateral  are  supple- 
mentary, the  other  two  are  supplementary  also. 

7.  What  is  the  size  of  the  obtuse  angle  formed  between  the 
bisectors  of  the  acute  angles  of  any  right  triangle  ? 

8.  Given  a  diagonal,  construct  the  corresponding  square. 

9.  Given    the    diagonals    of    a    rhombus,    construct    the 
rhombus. 

10.  A  so-called  T  joint  for  pipe  is  a  piece  made  in  the  form 
of  ^  a  letter  T  ;  the  angle  between  the 
arms  is  accurately  a  right  angle.  Show, 
by  §  52,  that  two  pipes  along  the  same 
wall  joined  to  the  same  main  pipe  by  T 
joints,  are  parallel. 

11.  Prove  that  the  bisector  of  the  exterior  angle 
at  the  vertex  of  an  isosceles  triangle  is  parallel  to 
the  base. 

Given  the  A  ABC  with  AC  =  BG  and  Z  DCE  = 
Z.BCE. 

To  prove  CE  \\  AB. 

12.  Prove  (using  Theorem  I),  that  in  an  equilateral  triangle 
the  bisector  of  any  angle  forms  two  congruent  triangles. 

13.  To  cut  two  converging  timbers  by  a  line 
AB  which  shall  make  equal  angles  with  them, 
a  carpenter  proceeds  as  follows :  Place  two 
squares  against  the  timbers,  as  shown  in  the 
figure,  so  that  AO==BO.  Show  that  AB  is 
then  the  required  line. 


ozrt 


EH 


I,  §  102]  MISCELLANEOUS  EXERCISES  85 

14.  Let  ABO  and  EST  be  two  congruent  triangles.  Prove 
that  the  medians  drawn  through  A  and  R  are  equal.  Prove  also 
that  the  altitudes  through  A  and  B  are  equal. 


C 


15.    In  the  figure,  CA  =  CB,  AD  =  BE. 
Prove  A  ADB  ^  A  ABE. 


D  ^E 

16.  In  a  right-angled  triangle,  if  one  of  the  acute  angles  is 
30°,  prove  that  the  side  opposite  is  half  the  hypotenuse. 

17.  Prove  that  the  bisectors  of  the  interior  angles  of  a  rec- 
tangle form  a  square. 

C 

18.  If  the  bisector  of  an  angle  of  a  triangle  is  / 
perpendicular  to  the  side  opposite  the  angle,  the  / 
triangle  is  isosceles.     Prove  this  statement.                 / 

ADB 

19.  Prove  that  the  line  joining  the  mid-points  of  the  non- 
parallel  sides  of  a  trapezoid  is  equal  to  half  the  sum  of  the 
bases. 

20.  i?  is  a  river,  and  it  is  required  to  find  the  distance  be- 
tween the  points  B  and  A  situated  on  the 
opposite  shores.     Show  that  this  may  be 
done  (without  crossing  the  stream)  as  fol- 
lows: 

(1)  Set  a  stake  at  some  point  C  in  line 
with  AB. 

(2)  Set  a  second  stake  at  some  point 
D  from  which  all  three  of  the  points  A,  B,  C  can  be  seen. 

(3)  Set  stake  at  E  in  line  with  DC  and  such  that  ED  =  DC, 
and  similarly  a  stake  at  F  which  shall  be  in  line  with  DB  and 
such  that  FD  =  DB. 

(4)  Set  stake  at  G  in  line  with  both  EF  and  AD. 
Then  FG  will  be  the  required  distance  AB. 


86 


RECTILINEAR  FIGURES 


[I,  §  102 


c              p 

D 

A 

2 

'                 B 

given 

21.  Prove  that  the  angle  between  the  bisectors  of  two  angles 
of  an  equilateral  triangle  is  double  the  third  angle. 

22.  In  the  triangle  KLN,  NM  is  per-  ^ 
pendicular  to  KL,  and  KM  =  MN  =  ML. 
Prove   that   KLN  is   an  isosceles   right 
triangle.                                                             K 

23.  Show  how  to  obtain  (as  a  crease) 
a  parallel  to  a  given  line  through  a 
given  point  on  a  piece  of  paper,  by- 
folding  the  paper,  and  prove  that  the 
result  is  correct. 

24.  A  man  wishes  to  measure  the  distance  between  two 
points  R  and  T  on  opposite  sides  of 
a  stream.  He  takes  a  line  VR  and 
measures  the  angle  TRV.  He  then 
walks  along  VR  prolonged  until  he 
reaches  a  point  S  where  Z  TSR  = 
\^TRV.     He  then  concludes  RS=RT.    Is  he  right  ?    Why  ? 

Note.  The  sailor  uses  this  principle  when  he  "  doubles  the 
angle  on  the  bow ''  to  find  his  distance 
from  any  object  on  shore.  Thus  if  he 
is  sailing  in  the  direction  ABC,  and  if 
Z  is  a  lighthouse,  he  measures  the 
angle  A,  and  if  he  notices  when  the 
angle  that  the  lighthouse  makes  with 
his  course  is  just  twice  the  angle  noted 
at  Af  then  BL  =  AB.  He  knows  AB  from  his  log;  hence  he 
knows  the  distance  BL. 

25.  Show  that  if  the  two  ends  of  one  side  of 
one  square  on  squared  paper  are  connected  by 
straight  lines  to  the  two  ends  of  any  parallel  side 
of  any  other  square  of  the  same  size,  taken  in  the 
same  order,  a  parallelogram  is  formed. 


D 

C 

/ 

/ 

/ 

/ 

/ 

/ 

1 

5 

5 

I,  §  102] 


MISCELLANEOUS  EXERCISES 


87 


■?rr".^.^Vrr.^r^.^rVr^- 


^^^^^^^^^^^^^J-^. 


26.  If  a  jointed  frame  of  the  form  of  a  parallelogram  is 
moved  so  that  the  angle  at  A  grows 
less,  show  that  (1)  the  angle  at  B  in- 
creases ;  (2)  the  diagonal  DB  decreases ; 
(3)  the  diagonal  AC  increases. 

27.  Can  the  frame  of  Ex.  26  be  braced 
effectively  by  flexible  wires  along  the  diagonals  ?  Why  does 
this  make  the  frame  quite  stiff?  This  principle  is  used  in 
bridge  building. 

28.  An  isosceles  trapezoid  is  one  whose  sides  (other  than  the 
bases)  are  equal.     Prove  that  the  diagonals 

of  an  isosceles  trapezoid  are  equal ;  also  that 
its  base  angles  are  equal. 

Given  the  trapezoid  ABCD  in  which  BC  = 
AD. 


A  F 


G  B 


To  prove  (1)  that  AO  =  BD  and  (2)  that  Z  ABC  =  Z  BAD. 
[Hint.     Draw  the  altitudes  DF,  CG,  and  provj  (2)  first.] 

29.  Show  that  if  a  rectangular  door  suspended  on  hinges 
hangs  out  of  a  vertical  line,  the  bottom  edge  of  door  is  out 
of  a  horizontal  line.  If  the  hinged  edge  leans  away  from 
the  vertical  line  by  2  in.  in  every  3  ft.  of  its  length,  show  that 
the  bottom  edge  rises  by  2  in.  in  every  3  ft.  of 
its  length. 

30.  A  quadrilateral  of  which  two  pairs  of  ad- 
jacent sides  are  equal  is  called  a  kite.  Show 
that  one  of  the  diagonals  divides  a  kite  into  two 
congruent  triangles ;  show  that  the  other  diago- 
nal divides  the  kite  into  two  triangles,  each  of 
which  is  isosceles. 

31.  Show  that  one  pair  of  opposite  angles  of  a  kite  are  equal. 
If  these  angles  are  both  right  angles,  show  that  the  other  two 
angles  are  supplementary. 


88  RECTILINEAR  FIGURES  [I,  §  102 

32.  Construct  a  triangle  having  given  the  mid-points  of  its 
sides.     See  Theorem  XXXII,  §  102. 

33.  What  is  the  locus  of  the  middle  points  of  all  straight 
lines  drawn  from  a  fixed  point  to  a  fixed  line  of  unlimited 
length? 

34.  Prove  that  if  the  diagonals  of  a  parallelogram  are  equal 
and  perpendicular  to  each  other,  the  figure  is  a  square. 

35.  Construct  an  equilateral  triangle,  having  given  its 
altitude. 

36.  If  in  A  ABC,  AC  =  BC,  and  if  AC 
is  extended  to  D  so  that  CD  =  AC,  prove 
that  DBA-AB.  Hence  show  how  to  draw 
a  perpendicular  to  a  line  AB  at  one  end 
B  without  extending  AB. 

37.  Show  how  to  trisect  (divide  into  three  equal  parts)  a 
straight  angle ;  a  right  angle ;  an  angle  of  45°. 

38.  ABC  is  a  right-angled  triangle.      BD  is  drawn  from 
the  right-angle  to  the  mid-point  of  the  hy- 
potenuse.    Prove  that  the  triangle  ABC  is 
thus  divided  into  two  isosceles  triangles. 

39.  Prove  that  the  sum  of  the  sides  of  a 
quadrilateral  is  greater  than  the  sum  of  its  diagonals,  but  less 
than  twice  their  sum. 

40.  Prove  that  the  difference  between  the  diagonals  of  a 
quadrilateral  is  less  than  the  sum  of  either  pair  of  opposite 
sides. 

41.  A  line  is  terminated  by  two  parallel  lines.  Through  its 
mid-point  any  line  is  drawn  terminated  by  the  parallels.  Prove 
that  the  second  line  is  bisected  by  the  first. 

42.  Prove  that  the  perpendiculars  drawn  from  the  extremities 
of  one  side  of  a  triangle  to  the  median  upon  that  side  are  equaL 


CHAPTER  II 


THE   CIRCLE 
PART  I.     CHORDS.     ARCS.     CENTRAL  ANGLES 

103.   Definitions.     The  circle  (§§  2,  23)  is  a  curve,  all  points 
of  which  are  equally  distant  from  a  point 
within,  called  the  center;  or  (§  99),  it  is 
the  locus  of  all  points  equally  distant  from 
a  given  point. 

By  the  definition  of  a  circle,  all  its  radii 
must  be  equal.    (See  §§2,  23.) 

Any   portion   of   the   circumference   is 
called  an  arc  (§  2). 

One  quarter  of  a  circumference  is  called 
a  quadrant. 

A  chord  is  a  straight  line  joining  the  extremities  of  an  arc. 

A  diameter  is  a  chord  that  passes  through  the  center. 

The  angle  between  any  two  radii  is  called  a  central  angle. 

In  Fig.  74,  the  central  angle  AOB  is  said  to  intercept  (cut  off) 
the  arc  AB  (written  AB  ) ;  while  the  arc  AB  is  said  to  subtend 
the  angle  AOB. 

An  area  bounded  by  two  radii  of  a  circle 
and  the  arc  between  them  is  called  a  sector. 

An  area  bounded  by  a  chord  of  a  ciicle 
and  its  arc  is  called  a  segment  of  the  circle. 

Two  circles  are  said  to  be  equal  when  the 
radius  of  the  one  is  equal  to  the  radius  of 
the  other. 

Two  circles  that  have  the  same  center  are  said  to  be  concentric 

89 


Fig.  75 


90 


THE  CIRCLE 


[II,  §  104 


104.  Postulates.  In  what  follows  we  shall  use  the  follow- 
ing facts  as  postulates : 

(1)  As  a  central  angle  increases,  its  intercepted  arc  increases, 
and  vice  versa  ;  and  as  a  central  angle  decreases,  its  intercepted 
arc  decreases,  and  vice  versa. 

(2)  In  the  saine  circle  {or  equal  circles),  equal  central  angles 
iiitercept  equal  arcs;  and  equal  arcs  subteyid  equal  central  angles. 


Fig.  76 

Thus,  in  Fig.  76,  the  equal  central  angles  AOB  and  A'O'B' 
intercept  the  equal  arcs  AB  and  A'B',  and  the  equal  arcs  AB 
and  A'B'  subtend  the  equal  central  angles  AOB  and  AOB'. 

105.  Rotation.  In  considering  the  relations  between  the 
angles  at  the  center  of  a  circle  and  their  intercepted  arcs,  it  is 
helpful  to  think  of  the  rotation  of  a  wheel  about  its  axle. 

During  such  a  rotation,  any  spoke  of  the  wheel  turns  through 
a  constantly  increasing  angle.  The  end  of  the  spoke  describes 
the  arc  of  the  circle  that  forms  the  rim  of 
the  wheel,  while  the  angle  described  by  the 
spoke  intercepts  on  the  rim  precisely  the 
arc  described  by  the  end  of  the  spoke. 

Thus,  any  two  spokes  of  the  same  wheel 
describe  equal  angles  in  equal  times.  The 
arcs  described  by  the  ends  of  the  two 
spokes  are  also  equal.     [(2),  §  104.] 

As  the  angle  a  spoke  describes  increases,  the  arc  that  its 
end  describes  also  increases,  that  is,  the  greater  of  two  angles 
at  the  center  intercepts  the  greater  arc.     [(1),  §  104.] 


II,  §  105]         CHORDS,  ARCS,  CENTRAL  ANGLES 


91 


£X£RCIS£S 

1.  What  is  the  central  angle  between  the  hands  of  a  clock 
when  it  is  three  o'clock  ?  Answer  the  same  question  for  four 
o'clock,  eight  o'clock,  and  half  past  nine. 

2.  Show  that  the  diameter  of  a  circle  is  its    q 
greatest  chord. 

[Hint.     In  the  figure,  AC  =  AO -\-0C  =  AO  +  OB. 
But  AO-h  OB>AB.;\ 


B 


0 


3.  Through  a  point  within  a  circle  to  construct  the  longest 
possible  chord. 

4.  Show  that  if  one  arc  of  a  circle  is  double  another  arc  of 
the  same  circle,  the  angle  at   the  center   sub-         p 
tended  by  the  first  is  double  that  subtended  by 
the  second. 


[Hint. 

§  104.] 


First  bisect  the  larger  angle  ;  then  apply  (2), 


5.  Show  that  if  one  arc  of  a  circle  is  n  times 
another  arc  of  the  same  circle,  the  angle  at  the  center  sub- 
tended by  the  first  is  n  times  that  subtended  by  the  second. 

6.  Show  that  if  a  circumference  is  divided  into  360  equal 
arcs,  the  central  angles  subtended  by  these  arcs  are  all  equal 
(one  degree) ;  show  that  the  number  of  degrees  in  any  central 
angle  is  equal  to  the  number  of  these  small  arcs  contained  in 
its  intercepted  arc. 

7.  Prove  that  two  intersecting  diameters  divide  a  circumfer- 
ence into  four  arcs  each  of  which  is 
equal  to  one  of  the  others. 

^  8.  The  diameter  AB  and  the  chord 
CD  are  prolonged  until  they  meet  at  E. 
Prove  that  EA  >  EC  and  EB  <  ED. 


92  THE  CIRCLE  [II,  §  106 

106.   Theorem  I.     In  the   same  circle   or  in    equal   circles, 
equal   arcs    subtend   equal 


chords.                                       /^ 

Given  the  two  equal  ©     / 
0  and  0'  in  which  j3=      I 

^ 

)k 

:t 

[O  is  the  symbol  for  circle; 
-^  is  the  symbol  for  arc.'] 

^         " 

Fig.  78 

^ 

To  prove  that  chord  AB  =  chord  A'B'. 

Proof.  Draw  the  radii  OA,  OB,  O'A',  0'B\  Then  in  the 
A  AOB,  A'O'B',  we  have 

OA  =  0A\  OB  =  O'B',  §  103 

and  ZO  =  Z  0'.  §  104 

Therefore  A  AOB  ^  A  A'O'B' ;  §  35 

hence  AB  =  A'B'. 

107.  Theorem  II.  (Converse  of  §  106.)  In  the  same  circle 
or  i7i  equal  circles,  equal  chords  subtend  equal  arcs. 

Given  the  two  equal  (D  0  and  0'  (Fig.  78)  in  which  the  chord 
AB  =  the  chord  A'B'. 

To  prove  that  AB  =  A'B'. 

Outline  of  proof.  Draw  the  radii  OA,  OB,  O'A',  O'B'.  Then 
show,  by  §  45,  that  A  AOB  ^  A  A' O'B'  and  apply  §  104. 

EXERCISES 

1.  Show  that  in  the  same  or  equal  circles  the  greater  of  two 
arcs  subtends  the  greater  chord,  and  vice  versa.  [Use  §  104  (1) 
and  §  76.] 

2.  Show  that  the  construction  of  an  angle  equal  to  a  riven 
angle  (§  7)  illustrates  §  107.  Hence  prove  that  the  construc- 
tion in  §  7  is  correct. 

3.  State  Theorems  I  and  II,  using  the  phrase  (a)  "  and  con- 
versely "  ;  (b)  "  and  vice  versa '' ;  (c)  ^'  if  and  only  if."    See  §  55. 


II,  §  108]        CHORDS,  ARCS,  CENTRAL  ANGLES 


93 


108.  Theorem  III.  A  diameter  perpendicular  to  a  chord 
bisects  the  chord  and  the  arc  subtended  by  it. 

Given   the   diameter  DF  ±  chord  AB 
at  K. 

To  prove  that  AK=  KB  and  that  AF  = 

FB. 

Proof.      Draw   the   radii   OA  and   OB 

Then 

6k=  ok,  and  OA=OB;      Why  ? 
hence     rt.  A  OKA  ^  rt.  A  0KB.      Why  ? 

Therefore 
AK=KB,  and  Z  AOK=Z.  BOK;  Why  ? 

hence  AF=  FB.  (2)  §  104 

Note  1.  In  the  figure  above  the  chord  AB  may  be  regarded 
as  subtending  not  only  the  arc  AFB,  but  also  the  larger  arc 
ADB.  It  is  customary  to  speak  of  the  two  as  tae  minor  arc 
and  the  major  arc  corresponding  to  the  given  chord.  Similarly, 
any  central  angle  subtends  both  a  minor  and  a  major  arc. 
Unless  otherwise  stated,  the  minor  arc  is  the  one  to  be  under- 
stood hereafter  in  any  statement  where  both  might  play  a  part. 

Note  2.  This  extremely  impoHant,  though  simple,  figure 
(Fig.  79),  occurs  in  the  greatest  variety  of  practical  affairs,  and 
in  many  geometric  theorems  and  constructions.  (See  §§  b,Q,  8, 
9,  40,  43,  44,  72, 100,  102,  and  Exs.  2,  p.  43;  1,  p.  45,  etc.) 


J 


EXERCISES 

1.  Prove  that  a  diameter  perpendicular  to  a  chord  bisects 
the  major  arc  subtended  by  it. 

2.  What  is  the  locus  of  the  mid-points  of  a  system  of  parallel 
chords  ? 

•*  3.  Prove  that  the  perpendicular  bisector  of  a  chord  passes 
through  the  center  of  the  circle  and  bisects  the  arcs  (major  and 
minor)  subtended  by  the  chord. 


94 


THE  CIRCLE 


[II,  §  109 


109.  Theorem  IV.  In  the  same  circle  or  in  equal  cir- 
cles, equal  chords  are  equally  distant  from  the  center; 
and,  conversely,  chords  that  are  equally  distant  from 
the  center  are  equal. 


Fig.  80 

Given  the  equal  chords  AB  and  A'B^  in  the  equal  (D  0  and  0'. 

To  prove   that  AB  and  A'B^  are  equally  distant  from  the 
centers  0  and  0',  respectively. 

Proof.     Draw  OD  J_  AB  and  O'D'  ±  A'B')  draw  also  the 
radii  OB  and  0'^'. 

Prove  that  A  DOB  ^  A  D'O'B', 

and  hence  OD  =  O'D'. 

In  the  converse,  it  is  given  that  the  chords  AB  and  A^B^  are 
equally  distant  from  the  centers  of  the  equal  ©  0  and  0'. 

To  prove  that  AB  =  A  'B'. 

Proof.     Show  that  A  DOB  ^  A  D'O'B' 
and  hence  that  DB  =  D'B'. 

But  AB  =  2DB  and  ^'5'  =  2  Z>'5' ;  §  108 

whence  AB:==A'B\  Why? 


EXERCISES 

1.  Show  that  two  boards  sawed  from  the  same  log,  or  from 
equal  logs,  at  equal  distances  from  the  center,  are  equal  in  width. 

2o  What  is  the  locus  of  the  mid-points  of  a  system  of  equal 
chords  in  a  circle  ? 


II,  §  111]        CHORDS,  ARCS,   CENTRAL  ANGLES 


95 


110.  Theorem  V,  In  the  same  circle,  or  in  equal  circles,  if 
two  unequal  chords  are  drawn,  the  longer  one  is  nearer  the 
center. 

Given  the  O  0  with  the  two  chords  AB 
and  CD  such  that  AB  >  CD.  Also,  let  OE 
and  Oi^be  the  perpendicular  distances  from 
0  to  AB  and  CD,  respectively. 

To  prove  that  OE  <  OF. 

Proof.  From  A  lay  off  the  chord  AB' 
=  CD.  Then  draw  the  perpendicular  OK, 
and  finally  join  K  to  E  by  the  line  KE. 

Since  AE^^AB  and  AK=  \  AB', 

it  follows  that       AE  >  AK,  for  AB  >  AB'. 

Hence  Zb  >  Z.c, 

and  consequently  Z  a  <  Z  d. 

Therefore  OE  <  OK. 

But  0F=  OK. 

Therefore  OE  <  OF. 


FiQ.  81 


§  108 
Given 
§  78 
Ax.  8 
§75 
§109 
Ax.  9 


111.  Corollary  1.  {Converse  of  §  110.)  In  the  same  circle 
or  in  equal  circles,  if  two  chords  are  unequally  distant  from  the 
center,  the  more  remote  is  the  less. 

EXERCISES 

1.  Show  that  a  board  sawed  from  a  circular  log  is  wider 
than  another  board  sawed  from  the  same  log  at  a  greater  dis- 
tance from  the  center. 

2.  Show  that  the  least  chord  that  can  be  drawn  through  a 
given  point  within  a  circle  is  the  chord  per- 
pendicular to  the  radius  through  that  point. 

[Hint.     Draw  any  other  chord  through  P  and  draw 
OC  perpendicular  to  it.    First  prove  0P>  OC.J 

3.  Show  that  a  diameter  is  longer  than  any 
other  chord. 


96 


THE  CIRCLE 


[II,  §  112 


Fig.  82 


112.  Chords.  The  relations  between  chords,  arcs,  and  cen- 
tral angles  of  the  same 

circle    appear   vividly  **^    D 

in  connection  with  ro- 
tation (§  105).  Thus 
the  chord  that  sub- 
tends the  arc  grows  as 
the  angle  grows ;  that 
is,  as  the  wheel  rotates, 
until  the  chord  reaches 
its  greatest  possible 
size,  the  diameter  of 
the  circle.  At  this 
time,  the  angle  the 
spoke  has  described  is 
180°,  or  a  straight 
angle.  As  the  rota- 
tion goes  on,  the  chord 

shrinks  again  to  less  than  the  length  of  the  diameter,  until  the 
wheel  has  made  one  complete  revolution,  when  the  chord  has 
shrunk  to  zero. 

If  a  circle  of  known  radius  is  drawn,  any  central  angle 
subtends  a  chord  of  some  definite  length.  Taking  the  radius 
as  1  unit,  the  lengths  of  the  chords  corresponding  to  various 
central  angles  for  every  degree  from  0°  to  90°  are  given  in  the 
table  of  chords,  Tables,  pp.  iii-vii. 

By  means  of  this  table,  any  angle  can  be  laid  off  from  any 
point  as  vertex,  by  drawing  a  circle  of  unit  radius  about  it. 

Thus,  the  chord  that  corresponds  to  a  central  angle  of  29""  is  very 
nearly  equal  to  \.  To  lay  off  29°  at  a  point  P  on  a  line  MN^  draw  a  circle 
X  of  unit  radius  about  P  as  center,  cutting  MN  at  §,  as  in  §  7,  p.  6.  Then 
draw  an  arc  y  about  Q  as  center  with  radius  \.  If  x  and  y  intersect  at  0, 
Z  OFQ  =  29°,  approximately. 

Let  the  student  lay  off  the  angles  of  31°,  69°,  etc.,  in  a  similar 
manner,  using  the  Tables. 


II,  §  113]        CHORDS,  ARCS,  CENTRAL  ANGLES  97 

113.  Angular  Speed.  The  facts  about  the  rotation  of 
wheels  or  other  parts  of  machinery  are  often  clearly  expressed 
in  terms  of  the  speed  with  which  the  wheel  (or  other  part)  is 
rotating.  This  speed  of  rotation  means  the  amount  of  angle 
through  which  the  wheel  turns  in  one  unit  of  time ;  for 
example,  we  say  that  a  certain  wheel  is  turning  at  the  rate 
of  four  revolutions  per  minute,  or  that  some  other  wheel  is 
rotating  50°  per  second.  Such  a  speed  of  rotation  of  a  wheel 
is  called  its  angular  speed. 

Which  of  the  two  wheels  just  mentioned  is  moving  the  faster  ?  The 
answer  to  this  question  is  found  by  changing  revolutions  per  minute  into 
degrees  per  second  as  follows :  The  first  wheel  is  rotating  four  revolu- 
tions per  minute.  Since  four  revolutions  means  1440^,  that  wheel  is 
making  1440°  per  minute,  or  1440°  every  60  seconds.  Dividing  by  60,  we 
find  that  it  is  going  24°  each  second.  Hence  the  second  wheel  is  rotating 
over  twice  as  fast. 

EXERCISES 

1.  Prove  that  in  the  same  circle,  or  in  equal  circles,  equal 
chords  subtend  equal  central  angles,  and  vice  versa. 

2.  Prove  that  the  greater  of  two  chords  of  a  circle  subtends 
the  greater  central  angle,  provided  the  central  angle  less  than 
180°  is  understood  in  each  instance. 

3.  Draw  a  circle  of  radius  one  inch.  Measure  approximately 
the  length  of  the  chord  of  each  of  the  following  angles :  45°, 
60°,  75°,  90°,  120°,  180°,  210°.  Check,  when  possible,  by  means 
of  the  tables,  pp.  iii-vii. 

4.  If  an  angle  is  doubled,  is  its  chord  doubled?  Compare 
the  lengths  of  chords  of  angles  of  45°  and  90°.  Compare  the 
lengths  of  chords  of  angles  of  60°  and  180°.  Check  your  results 
by  means  of  the  tables  on  pp.  iii-vii. 

5.  If  one  wheel  is  rotating  through  5  revolutions  per  minute, 
and  another  through  30°  per  second,  which  is  rotating  the  more 
rapidly  ? 


98 


THE   CIRCLE 


[11,  §  113 


MISCELLANEOUS   EXERCISES.     PART  I 

1.  Show  that  a  chord  equal  to  the  radius 
subtends  an  angle  of  60°. 

2.  If  a  circle  is  drawn  through  all  four 
vertices  of  a  square,  show  that  the  arcs  in- 
tercepted by  its  sides  are  equal.  Prove  that 
each  arc  subtends  a  central  angle  of  90°. 

3.  Show  that  if  the  vertices  of  an  equilateral  triangle  all  lie 
on  a  circle,  each  side  intercepts  an  arc  of  120°. 

4.  Show  that  if  the  vertices  of  an  equilateral  hexagon  all 
lie  on  a  circle,  each  side  intercepts  an  arc  of  60°.  Show  how 
to  construct  such  a  figure. 

5.  If  a  wheel  is  rotating  two  revolutions  per  minute,  what 
part  of  a  revolution  does  it  make  in  one  second  ?  How  many 
degrees  does  it  turn  through  in  one  second  ? 

6.  The  earth  revolves  once  in  24  hours.  How  many  degrees 
correspond  to  one  hour  ?  Ans.  15°. 

7.  A  jointed  extension  rod  —  such  as  that  used  on  desk  lights 
—  is  made  by  having  a  piece  in  the  form  of  a  circular  ring  fit 
over  the  end  of  another  circular  piece,  in  the  manner  illustrated 


in  the  figure.    An  excessive  motion  is  prevented  by  projections 
at  the  points  B  and  R'  and  at  /S  and  S'. 

Show  that  when  the  stop  S  has  reached  E,  the  line  AB  has 
turned  away  from  the  line  DC  by  half  the  difference  between 
the  angles  BOS  and  EOF. 


n,§ii4] 


TANGENTS  AND  SECANTS 


99 


PART   II.     TANGENTS   AND   SECANTS 


CIRCUMSCRIBED  AND   INSCRIBED   TRIANGLES 

114.   Definitions.     A  line  of  indefinite  length  wliich  cuts  a 
circle  is  called  a  secant. 

A  line  of  indefinite  length  which  touches 
a  circle  in  but  one  point  is  called  a  tangent; 
this  point  P  is  then  called  the  point  of  con- 
tact, or  point  of  tangency. 

A  triangle  or  other  polygon  is  said  to 
be  inscribed  in  a  circle  when  its  vertices 
all  lie  on  the  circumference.  Under  the 
same  conditions,  the  circle  is  said  to  circumscribe  the  polygon. 


Fig.  8;i 


Inscribed  Triangle  Circumscribed  Triangle 

Fig.  84 


A  triangle  or  other  polygon  is  said  to  be  circumscribed  about 
a  circle  when  its  sides  are  all  tangent  to  the  circle.  Under 
the  same  conditions,  the  circle  is  said  to  be  inscribed  in  the 
polygon. 

A  good  idea  of  a  line  tangent  to  a  circle  is  obtained  by  placing 
a  coin  (representing  a  circle)  against  the  edge  of  a  ruler  (rep- 
resenting a  straight  line). 

A  tangent  to  a  circle  may  be  roughly  drawn  by  placing  a 
ruler  so  that  its  edge  just  meets,  but  does  not  cut  across,  the 
circumference.  If  several  tangents  are  drawn  to  the  same 
circle  in  this  manner,  and  extended  to  meet,  a  circumscribed 
polygon  results. 


100 


THE  CIRCLE 


[II,  §  115 


115.   Theorem  VI.    A  line  perpendicular  to  a  radius  at  its 
extremity  is  tangent  to  the  circle. 


Given  the  O  0  and  the  line  AB  ±  to  the  radius  OE  at  its 
extremity. 

To  prove  that  AB  is  a  tangent  to  the  circle. 

Proof.     Take  any  point  D  on  AB  except  E  and  draw  OD. 

Then  OD  >  OE.  §  77 

Therefore  the  point  D  is  not  on  the  circle ;  §  103 

that  is,  no  point  of  AB  except  E  lies  on  the  circle. 

Whence  AB  is  a  tangent.  §  114 

116.  Corollary  1.  A  tangent  to  a  circle  is  perpendicular  to 
the  radius  drawn  to  the  point  of  contact. 

[Note  that  we  know  that  OjE'<  OD.     Then  apply  §  77.] 

117.  Corollary  2.  A  perpendicular  to  a  tangent  at  its  point 
of  contact  passes  through  the  center  of  the  circle. 

[Hint.  Draw  the  radius  to  the 
point  of  contact  and  apply  Corollary 
1,  together  with  §  58.] 

118.  Theorem    VII.      Two 

tangents  drawn  to  a  circle  from 
a  point  outside  are  of  equal 
length. 

[The  proof  is  left  to  the  student, 
with  the  aid  of  Fig.  86.] 


Fig.  86 


n,§ii8] 


TANGENTS  AND  SECANTS 


101 


£X£RCISES 

1.  Prove  that  two  tangents  drawn  to  a  circle  at  the  extremi- 
ties of  a  diameter  are  parallel. 

2.  Construct  a  tangent  to  a  circle  parallel  to  a  given  line. 

3.  Draw  two  concentric  circles  (§  103)  and  prove  that  all 
chords  of  the  greater  circle  that  are  tangent  to  the  smaller 
circle  are  equal. 

4.  Let  0  and  0'  be  two  circles  which  are  tangent  to  each 
other ;  that  is,  which  have  but  one  point  in  common.     Prove 

(1)  that  the  line  joining  0  and  0'  (line   of  centers)  passes 
through  the  point  common  to  the  two  circles  (see  12,  §  31) ;  and 

(2)  that  a  perpendicular  to  the  line  OO  at  the  common  point 


P  will  be  a  tangent  to  both  circles.  Consider  the  case  in 
which  the  circles  are  tangent  externally,  and  also  the  case  in 
which  they  are  tangent  internally,  as  indicated  in  the  figures. 

6.  A  railroad  curve  joining 
two  pieces  of  straight  track  AB 
and  CD,  is  usually  a  circular 
arc  tangent  at  B  and  C  to  AB 
and  CD,  respectively.  Show 
that  the  center  O  of  the  cir- 
cular arc  is  the  intersection  of 
the  perpendiculars- to  AB  and  CD  at  B  and  C,  respectively. 

6.  If  AB  and  CD  (Ex.  5)  are  extended  to  meet  at  E,  show 
that  EO  bisects  the  angle  at  0.  Hence  show  that  F,  the 
intersection  of  EO  and  BC,  is  the  center  of  Sb. 

7.  Show  that  the  two  radii  and  the  tangents  at  their  ex- 
tremities (Fig.  86,  or  figure  for  Ex.  5)  form  a  kite  ;  that  is 
(Ex.  30,  p.  87),  two  pairs  of  its  adjacent  sides  are  equal 


102 


THE  CIRCLE 


[II,  §119 


119.  Theorem  VIII.     Tivo  parallel  lines  intercept  equal 
arcs  on  a  circle. 


Case  I.     WJien  the  parallels  are  a  tangent  and  a  secant. 

Given  the  tangent  AB  II  the  secant  EF  (Fig.  87  a) ;  also,  let 
C  be  the  point  of  contact  of  AB. 

To  prove  that  EQ=6f. 

Proof.  Draw  the  diameter  CD.  Then  CD1.AB',  §  116 
hence  OD  _L  EF,  Why  ? 

and  therefore  EC  =  CF.  §  108 

Case  II.      When  the  parallels  are  both  secants. 

Given  the  parallel  secants  AB  and  EF  (Fig.  87  b). 

To  prove  BF=  AE. 

Proof.  Draw  DC  II  AB  and  tangent  to  the  circle.  Let  Jf  be 
the  point  of  contact. 

Theni>CMI^F.  Why? 

Whence  FM=  EM,  and  BM=  AM.  Case  I 

Therefore  BF=  AE.  Ax.  2 

Case  III.     When  the  parallels  are  both  tangents. 
Given  AB  and  EF  parallel  tangents  touching  the  circle  at 
M  and  K,  respectively  (Fig.  87  c). 

To  prove  that  MLK=  MRK. 

Outline  of  proof.  Draw  CD  II  AB  (Fig.  87  c) ;  and  prove 
MC=Mb,  and  KC  =  KD,     Then  apply  Ax.  1, 


n,  §  119] 


TANGENTS  AND  SECANTS 


103 


EXERCISES 

1.  Show  that  a  tangent  parallel  to  any  chord  of  a  circle  has 
its  i^oint  of  tangency  in  the  center  of  the  arc  that  the  chord 
intercepts. 

2.  State  and  prove  the  converse  of  Theorem  VITI,  §  119. 

[Hint.     (Case  I)  :  Given  GF=  CE\  \jo  prove  AB II  EF. 

Draw  the  diameter  CD,  also  the  radii  OE,  OF. 
Then  CD  ±  AB.  '  (Why  ?) 

Now,  since  CF  =  CE,  we  have  Z  COF  =  Z  COE. 

(Why  ?) 

Therefore  A  FGO  ^  A  EGO  (Why?),  so  that 
ZFGO  =  ZEGO. 

Therefore  CD±EF.  Since  CD±AB  and  CD 
±  EF,  we  have  ABW  EF  as  desired.] 

3.  If  every  vertex  of  a  trapezoid  lies  on  a 
circle,  prove  that  it  is  isosceles;  that  is,  that 
BC=AD. 

4.  If  a  thick  board  is  sawed  out  of  a  circular 
log,  the  circular  arc  that  forms  one  edge  is  equal 
to  that  which  forms  the  other  edge.     Connect  this  fact  with 
§  119. 

5.  Let  ABC  be  an  isosceles  triangle,  with  Z.  B  =  Z  C;  and 
let  AD  be  its  altitude  from  A  to  BC.      Draw 
a  circle  with  center  at  A  and  with  radius  AD, 
cutting  the  sides  ^B  and  AC  in  F  and  J5J,  respec- 
tively.    Show  that  EF  II  BC. 

6.  Draw  a  secant  intersecting  two  concentric 
circles  and  prove  that  the  portions  intercepted  between  the  two 
circles  are  equal. 

7.  Prove  that  if  every  vertex  of  a  parallelogram  lies  on  a 
circle,  any  two  opposite  sides  are  equidistant  from  the  center. 

8.  Prove  that  if  a  polygon  is  inscribed  in  a  circle,  the  per- 
pendicular bisectors  of  the  sides  meet  in  a  point. 


104  THE  CIRCLE  [II,  §  120 

120.  Theorem  IX.  Through  three  given  points  not 
all  on  the  same  straight  line,  one  and  only  one  circle 
can  he  draivn. 


Given  the  three  points  A,  B,  and  0  not  all  on  the  same 
straight  line. 

To  prove  that  one  and  only  one  circle  can  be  passed  through 
A,  B,  and  C. 

Proof.  The  locus  of  all  points  equally  distant  from  A  and  B 
is  the  perpendicular  bisector  DE  of  the  line  AB,  while  the  locus 
of  all  points  equally  distant  from  B  and  C  is  the  perpendicular 
bisector  FG  of  the  line  BO.  §  100 

Therefore  the  intersection  0  of  DE  and  FG  is  equally  dis- 
tant from  all  three  of  the  points  A,  B,  and  C. 

Hence,  the  circle  drawn  with  0  as  center  and  with  a  radius 
equal  to  the  line  AO  will  pass  through  A,  B,  and  C. 

That  this  is  the  only  such  circle  follows  from  the  fact  that 
the  lines  DE  and  FG  can  intersect  in  but  one  point.       4,  §  31 

Note.  Theorem  IX  is  frequently  stated  in  the  following 
brief  form  :  Three  points  determine  {fix)  a  circle.  The  proof 
also  shows  how  to  construct  the  circle  passing  through  three  given 
points.  Thus,  given  A,  B,  and  G  (Fig.  88);  draw  AB  and  BC 
and  erect  their  perpendicular  bisectors,  DE  and  FG.  The 
intersection  0  of  DE  and  FG  is  the  center  of  the  desired 
circle ;  its  radius  is  one  of  the  equal  distances  OA,  OB,  or  00. 


II,  §  123]  TANGENTS  AND  SECANTS  105 

121.  Corollary  1.  A  cirde  may  be  dravjn  to  circumscribe  any 
tnangle. 

[Draw  the  three  perpendicular  bisectors  of  the  sides  of  the  triangle.] 

122.  Corollary  2.  TTie  perpendicular  bisectors  of  the  sides  of 
a  triangle  meet  in  a  point.     (Compare  Theorem  XXX,  §  102.) 

This  point  is  called  the  circumcenter,  because  it  is  the  center 
of  the  circumscribed  circle. 

123.  Corollary  3.  A  circle  may  be  completed  if  any  arc  of  it 
is  given, 

[Hint.    Take  three  points  on  that  arc  and  draw  a  circle  through  them.] 

EXERCISES 

1.  Try  to  prove  Theorem  IX,  §  120,  for  three  points  A,  B, 
C,  that  lie  on  a  straight  line.  At  what  place  does  the  proof 
break  down,  and  why  ? 

2.  Draw  a  triangle  of  any  shape  and  then  construct  the 
circle  which  passes  through  its  vertices. 

3.  How  many  circles  can  be  drawn  through  two  given 
points  ?     What  is  the  locus  of  the  centers  of  all  such  circles  ? 

4.  How  many  circles  having  a  given  roilius  can  be  drawn 
through  two  given  points  ?  Is  it  always  possible  to  have  one 
such  circle  ?     Why  ? 

5.  Show  how  to  construct  the  center  of  a  given  circle. 
[Hint.     Erect  the  perpendicular  bisectors  of  any  two  chords.] 

6.  Show  that  a  circle  can  be  circumscribed  about  a  given 
quadrilateral  if  (and  only  if)  the  perpendicular  bisectors  of  the 
sides  all  meet  in  a  single  point. 

7.  Show  that  a  circle  may  be  circumscribed  about  any 
square ;  any  rectangle. 

8.  Show  that  the  perpendicular  bisectors  of  the  opposite 
sides  of  a  parallelogram  are  parallel  unless  they  coincide. 
Hence  show  that  an  inscribed  parallelogram  is  a  rectangle. 


106  THE  CIRCLE  [II,  §  124 

124.  Theorem  X.  A  circle  may  he  inscribed  in  any 
triangle.  C 

Given  the  triangle  ABC.  ^  y^j\ 

To  prove  that  a  circle  may  be  y/\~i\- 

inscribed  in  it.  ^/f    "^^je^    rva 

Proof.     Draw  the  bisectors  of  >'''''0^'\^''^^^~>C  \ 

the  angles  A,  B,  C:  these  three     .  j^^^       \^  I  ^^^^A  p 
bisectors  meet  in  a  point  I.    §  102  ^ 

Draw  the  perpendiculars  /L, 
IM,  INy  to  the  three  sides  a,  b,  c,  respectively ;  then 

IL  =  IM=m.  §101 

Hence  a  circle  with  radius  r  =  IL  is  tangent  to  a,  b,  and  c 
at  Lj  Mj  and  2^,  respectively.  §  115 

Note.  The  preceding  proof  involves  also  the  proof  of  the 
construction  of  the  inscribed  circle  in  any  triangle.  Compare  §  120. 

125.  Corollary  1.  A  circle  drawn  from  any  point  on  the 
bisector  of  an  angle,  ivith  a  radius  equal  to  the  distance  from  that 
point  to  one  side,  is  tangeyit  to  both  sides  of  the  angle. 

EXERCISES 

1.  How  many  circles  can  be  drawn  tangent  to  each  of  two 
given  intersecting  lines  ?     What  is  the  locus  of  their  centers  ? 

2.  What  is  the  locus  of  the  centers  of  the  circles  of  Ex.  1, 
in  case  the  two  given  lines  are  parallel  ? 

3.  Show  that  a  circle  can  be  inscribed  in  a  given  quadrilat- 
eral if  the  bisectors  of  the  four  angles  all  meet  in  a  single  point. 

4.  Show  that  a  circle  can  be  inscribed  in  any  square. 

5.  Show  how  to  round  off  the  vertices  of  any  triangle  by 
circular  arcs.  (See  §  125.)  This  process  is  used  in  rounding 
off  the  corners  of  triangles  that  occur  in  a  triangular-shaped 
piece  of  ground,  so  as  to  build  a  fence  or  a  sidewalk  or  a  build- 
ing without  sharp  corners.     (See  also  Ex.  5,  p.  101.) 


II,  §  127]  MEASUREMENT  OF  ANGLES  107 

PART   III.     MEASUREMENT   OF   ANGLES 

126.  Numerical  Measure.  In  order  to  measure  any  quantity, 
say  a  line  of  fixed  length,  we  must  first  select  the  unit  which 
we  are  to  use.  In  the  case  of  a  fixed  line,  the  customary  unit 
would  be  either  the  inch,  the  foot,  the  centimeter,  the  yard,  or 
any  one  of  several  others.  In  the  case  of  an  area,  the  unit 
might  be  a  square  inch  or  a  square  foot  or  an  acre,  or  any  one 
of  several  others. 

Having  once  selected  our  unit,  the  process  of  measuring 
consists  in  obtaining  some  idea  of  the  relative  size  of  the 
given  quantity  as  compared  to  that  of  the  chosen  unit. 
Thus,  in  the  case  of  the  fixed  line,  we  lay  a  yardstick  along- 
side of  it,  and  read  off  by  means  of  the  scale  provided  for  the 
purpose,  a  number  which,  at  least  with  some  degree  of  ac- 
curacy, tells  us  the  relative  size  of  the  line  in  question  to 
that  of  the  inch. 

Even  though  we  cannot  usually  determine  in  this  way  the 
exact  length  of  a  line,  owing  to  imperfections  both  in  our  in- 
struments and  our  eyesight,  still  we  suppose,  and  it  is  indeed 
an  axiom  of  measurement,  that  there  always  exists  in  every  case 
just  one  number  which  does  express  exactly  the  length  in  terms  of 
the  unit.  This  number  is  called  the  numerical  measure  of  the 
given  line,  corresponding  to  the  unit  selected. 

In  general,  the  numerical  measure  of  any  quantity  of  any 
kind  is  a  number,  obtained  as  above,  which  expresses  the 
relative  size  of  the  quantity  to  some  unit  of  the  same  kind 
selected  in  advance. 

127.  Ratio.  The  numerical  measure  of  one  quantity  divided 
by  the  numerical  measure  of  a  second  quantity  of  the  same 
kind,  provided  the  same  unit  has  been  used  in  each  case,  is 
called  the  ratio  of  the  first  quantity  to  the  second.  Thus,  the 
ratio  of  6  feet  to  8  feet  is  6/8  or  3/4;  again,  the  ratio  of 
6  inches  to  1  yard  is  6/36  or  1/6. 


108  THE  CIRCLE  [II,  §  128 

128.   Commensurable    and    Incommensurable  Quantities. 

Let  AB  and  CD  be  two  straight  lines  of  different  length  and 
let   it  be  supposed  that  a  certain  unit    ^,     ,     ,     ,     .^ 

of  length,  as  MN,  is  contained  an  exact 

(integral)  number  of  times  (that  is,  with-      '     '     '     '     '     ^ 
out  any  remainder)  in  AB.    For  example,  M— ^N 

let  MN  be  contained  4  times  in  AB.  ^^"-  ^ 

Then  MJSF  may  also  be  contained  an  exact  number  of  times 
in  the  other  line  CD,  but  more  often  this  will  not  be  the 
case;  ordinarily  the  unit  JOT"  will  be  contained  in  CD  a  certain 
exact  number  of  times  plus  a  remainder  x,  which  will  be  less 
than  MJSr.  This  is  illustrated  in  Fig.  90,  in  which  MN  is  con- 
tained 5  times  in  CD,  with  a  remainder  x. 

Now,  if  we  select  a  very  small  unit  MN  that  is  contained  an 
exact  number  of  times  in  AB  as  before,  we  obtain  a  very  small 
remainder  x  when  the  same  unit  is  applied  to  CD.  However, 
it  may  happen  that  we  can  never  take  MN  so  small  that  the 
corresponding  remainder  x  will  turn  out  to  be  exactly  zero. 
In  this  case  the  two  lines  AB  and  CD  (Fig.  90)  are  said  to 
be  incommensurable. 

Examples  of  incommensurable  lines  occur  frequently  in 
geometry.  Thus,  it  will  be  seen  later  that  in  any  ^ 
isosceles  right  triangle  a  side  and  the  hypotenuse 
constitute  two  incommensurable  lines;  that  is,  in 
Fig.  91,  AB  and  BC  are  incommensurable.  BC 
may  be  thought  of  as  the  diagonal  of  a  square  of 
which  AB  and  AC  are  two  sides.  ^^^'  ^^ 

Two  other  interesting  incommensurable  quantities  are  the 
diameter  of  any  circle  and  the  length  of  its  circumference. 
These  will  be  considered  in  detail  in  Chapter  Y. 

If,  on  the  other  hand,  it  is  possible  to  choose  a  unit  which  is 
contained  an  exact  number  of  times  in  AB  (Fig.  90),  and  also 
an  exact  number  of  times  in  CD,  then  the  lines  AB  and  CD 
are  said  to  be  commensurable. 


IT,  §  129]  MEASUREMENT  OF  ANGLES  109 

129.  Other  Cases.  Limits.  Thus  far  we  have  spoken  only 
of  commensurable  and  incommensurable  lines,  but  similar 
definitions  apply  to  angles,  arcs,  or  any  other  sorts  of  quantities. 
Thus,  two  angles  are  commensurable  when  a  sufficiently  small 
unit  will  be  contained  in  each  an  exact  number  of  times,  and 
they  are  incommensurable  when  no  unit  which  is  contained  an 
exact  number  of  times  in  the  one,  will  at  the  same  time  be 
contained  an  exact  number  of  times  in  the  other. 

In  any  case,  the  remainder  x  can  be  made  as  nearly  equal  to 
zero  as  we  please  by  taking  the  unit  sufficiently  small :  we 
often  say  that  x  may  be  made  to  approach  zero. 

If  a  variable  quantity  approaches  a  fixed  quantity  as  nearly  as  we 
please,  the  fixed  quantity  is  called  the  limit  of  the  variable  one.  Thus,  in 
the  preceding  paragraph,  we  might  say  that  the  limit  of  x  is  zero. 

EXERCISES 

1.  State  accurately  what  is  meant  by  two  incommensurable 
arcs ;  two  incommensurable  areas. 

2.  Show  that  the  remainder  a;  of  §  128  can  always  be  made 
less  than  \  in. 

3.  Show  that  the  remainder  a;  of  §  128  can  be  made  smaller 
than  one  thousandth  of  an  inch ;  one  millionth  of  an  inch. 
Hence  show  that  any  two  lengths  can  be  measured  in  terms 
of  a  common  (small)  unit,  except  for  a  remainder  that  is  less 
than  the  human  eye  can  see. 

4.  What  is  the  smallest  unit  shown  on  your  ruler  ?  Can  you 
see  lengths  less  than  this  smallest  division  ?  If  a  line  whose 
length  you  are  measuring  is  not  an  exact  number  of  these 
smallest  units,  how  can  you  estimate  its  exact  length  ? 

5.  Estimate  the  exact  width  of  one  line  of  type  on  this  page. 

6.  The  length  Of  a  material  object  will  change  on  account 
of  expansion  due  to  changes  in  temperature.  Show  that  a  unit 
can  be  chosen  so  small  that  the  remainder  a;  (§  128)  will  be  less 
than  the  expansion  due  to  a  temperature  change  of  one  degree. 


110 


THE   CIRCLE 


[II,  §  130 


130.  Theorem  XI.    In  the  same  circle^  or  in  equal  cir- 
cles,    two    central 
angles  have  the  same 
ratio   as    their   inter- 
cepted arcs. 

Given  the  two  equal 
circles  O  and  0' ;  also 
let  AOB  be  any  central 
angle  in  0,  and  let  A'O'B'  be  any  central  angle  in  0'. 

To  prove  that  the  ratio  of  Z.  AOB  to  Z.  A'O'B'  is  the  same 
as  that  of  arc  AB  to  arc  A'B'. 

Proof,     (a)    When  arc  AB  and  arc  A'B'  are  commensurable. 

In  this  case  a  unit  of  arc  may  be  found  (see  §  128)  which 
is  contained  an  exact  number  of  times  in  both  the  arc  AB  and 
the  arc  A'B'.  (Compare  Ex.  5,  p.  91.)  Let  m  be  such  a  unit, 
and  let  us  suppose  that  the  number  of  times  it  is  contained  in 
arc  AB  is  r,  while  the  number  of  times  it  is  contained  in  arc 
A'B'  is  s.     Then 

a.YG  AB  _r 
s 


(1) 


arc  A'B' 

Now  divide  the  arc  AB  into  its  r  divisions,  each  of  length 
771,  and  through  the  points  of  division  draw  radii  to  the  center 
0.  Likewise,  divide  the  arc  A'B'  into  its  s  divisions,  each  of 
the  same  length  m,  and  draw  radii  through  the  points  of  divi- 
sion to  the  center  0'.  Then  Z  AOB  is  divided  into  r  equal 
angles,  while  Z  A'O'B'  is  divided  into  s  equal  angles  of  the 
same  size  (§  104)  ;  therefore 

ZAOB^r 
s 


(2) 


Z  A'O'B' 
Erom  (1)  and  (2)  it  follows  that, 

ZAOB       SiTcAB 


Z  A'O'B'     ^ic  A'B' 


II,  §  130] 


MEASUREMENT  OF  ANGLES 


111 


It  would  remain  to  prove  Theorem  XI  when  arc  AC  and  arc 
A'C  are  incommensurable  ;  but  this  proof  is  interesting  only 
theoretically.  Instead  of 
giving  a  proof,  we  may 
assume  as  a  postulate 
that  if  any  two  geomet- 
ric ratios  are  equal  whenr 
ever  their  terms  are  com- 
mensurable,  they  are  equal 
also  when  their  terms  are 
incommensurable. 

The  following  proof  may  then  be  omitted  at  the  discretion  of 
the  teacher : 

Proof.     (6)    When  arc  AB  and  arc  A'B'  are  incommensurable. 

In  this  case  if  we  take  any  unit  of  arc  m  which  is  contained  an  exact 
number  of  times  in  the  arc  AB  and  apply  it  to  the  arc  A'B',  there  will  re- 
main after  the  last  point  of  division  a  certain  arc  B"B'  less  than  m.    §  128 

But  whatever  the  choice  of  m,  we  shall  have 
ZAOB         SLTC  AB 


Fig.  93 


(3) 


Case  (a) 


ZA'O'Bf'  Arc  A'B" 
Now,  as  m  is  taken  smaller  and  smaller,  this  equation  (3)  continues 
true  at  every  step.  At  the  same  time,  the  individual  membere  of  the  same 
equation  are  changing,  but  only  to  the  extent  that  ZA'O'B"  comes  closer 
and  closer  to  ZA'O'B',  while  arc  A'B"  comes  closer  and  closer  to  A'B'. 
Thus,  by  taking  m  sufficiently  small,  we  can  bring  the  first  and  second 
members  of  (3)  as  near  as  we  please  to  the  respective  values 
,^x  ZAOB        arc  AB 

^  ^  ZA'O'B''    SiYcA'B'' 

These  last  ratios,  then,  differ  by  as  little  as  we  please  from  the  equal 
ratios  in  (3)  ;  hence  they  themselves  differ  from  each  other  by  as  little  as 
we  please.  But  this  is  the  same  as  saying  that  these  ratios  (4)  are  actually 
equal,  for,  if  they  were  unequal  the  difference  between  them  could  not  by 
any  method  of  reasoning  be  shown  to  be  as  small  as  we  please,  but  would 
always  remain  greater  than  some  definite  amount ;  that  is,  in  fact,  what 
unequal  means.     Hence, 

ZAOB   _  arc  AB 
ZA'O'B'     axe  A'B'' 


112 


THE  CIRCLE 


[II,  §  131 


Note.  Theorem  XI  shows  that  the  number  of  angulai 
units  in  a  central  angle  is  equal  to  the  number  of  units  of  arc 
which  the  angle  intercepts,  if  a  unit  of  arc  subtends  a  unit 
angle.  Thus,  the  number  of  degrees  in  any  central  angle  is 
the  same  as  the  number  of  degrees  in  the  arc  it  ^ 

intercepts.    This  fact  is  expressed  by  saying  that 
A  central  angle  is  measured  by  its  intercepted  arc. 

131.  Definition.  An  angle  formed  by  the 
intersection  of  two  chords  in  the  circumfer- 
ence of  a  circle  is  called  an  inscribed  angle ; 
or,  the  angle  is  said  to  be  inscribed  in  the  circle. 

132.  Theorem  XII.  An  inscribed  angle  is  measured  by  one 
half  of  its  intercepted  arc. 

B  B  B 


Fig.  95  (a) 


Fig.  95  (c) 


Given  the  inscribed  Z  ABC  intercepting  AO  in  the  circle  O. 
To  prove  that  Z  ABC  is  measured  by  one  half  of  the  arc  AC. 

Case  1.      When  the  center  of  the  circle  lies  on  one  of  the  sides 
of  the  angle,  as  AB.     (Fig.  95  a.) 

Proof.     Draw  DC ;  then  OC==OB; 


hence 

But 
therefore 

Since 
it  follows  that 


ZOBC=ZOCB. 

Z  OBC+  Z  OCB  =  Z  AOC) 

2ZABC:=ZA0C 

Z  AOCis  measured  by  AC, 

Z  ABC  is  measured  by  ^  AC. 


Why? 
Why? 
Why? 

Ax.  9 
Note,  §  130 

Ax.  4 


II,  §  135] 


MEASUREMENT   OF  ANGLES 


113 


Case  2.     ^VJlen  the  center  of  the  circle  lies  within  the  angle. 

Proof.     Draw  the  diameter  BD.     (Fig.  95  6.) 

Then  Z  ABD  is  measured  by  ^  AD,  Case  1 

and  Z  DBC  is  measured  by  |-D0;  Case  1 

hence  Z  ABD  4-Z  DBC  is  measured  by  |(AD  +  DC) ; 
that  is,  Z  ^J56'  is  measured  by  ^  ^C. 

Case  3.     Wheii  the  center  of  the  circle  lies  outside  the  angle. 

Proof.     Draw  the  diameter  BD.     (Fig.  95  c.) 

Then  Z  DBC  is  measured  by  \  DC,  Case  1 

and  Z  DJ5JL  is  measured  by  \  DA ; 

hence  Z  i)^0-  Z  Z)B^  is  measured  by  ^{DG-DA) ; 
that  is,         Z  ^5(7  is  measured  by  \  AC.  C. 

133.  Corollary  1.  Any  angle  inscribed 
in  a  semicircle  is  a  right  angle  ;  as  the  angle 
BCA  in  the  figure. 

134.  Corollary  2.  Any  angle  inscribed  in 
a  segment  (see  §  131)  greater  than  a  semicircle 
is  acute f  ivhile  any  angle  inscribed  in  a  segment 
less  than  a  semicircle  is  obtuse. 

Thus,  in  the  figure,  ABC  is  acute,  while 
AB'C  is  obtuse. 

135.  Corollary  3.  All  angles  inscribed  in 
the  same  segment  are  equal. 

Thus,  in  the  figure,  the  angles  ABiC, 
AB^Cj  AB^C,  are  all  equal. 

FiQ.  98 


114 


THE  CIRCLE 


[II,  §  135 


EXERCISES 

1.  A  thin  elastic  band  is  stretched  along  tha  diameter  AB 
of  a  circle  and  then  pinned  firmly  to  the  circumference  at  the 
two  points  A  and  B.    If  it  be  now  stretched 

aside  by  means  of  a  pencil  point  so  that 

it    takes   the   position   indicated   by   the 

dotted  line ;  that  is,  so  that  a  third  point 

C  of  the  band  lies  on  the  circumference, 

what  can  be  said  of  the  angle  ACB?     As 

the  pencil  is  allowed  to  move  about  the  circumference,  the 

band  meanwhile  sliding  over  the  pencil  point,  how  does  the 

angle  ACB  change  ? 

2.  A  corner  of  a  piece  of  cardboard  (of  the 
usual  rectangular  shape)  is  pressed  tightly 
against  two  pins  E  and  F  stuck  into  a  board 
below.  The  cardboard  is  now  turned  in  all 
possible  positions,  keeping  it  flat  against  the 
board.     What  is  the  locus  of  the  point  A  ? 

3.  Prove  by  means  of  §  132  that  the  sum  of  the  three  angles 
of  a  triangle  is  two  right  angles. 

4.  If  a  circumference  be  divided  into  four  equal  arcs 
(quadrants),  show  that  the  chords  which  join  the  extremities 
form  a  square. 

5.  Show  by  §  132  that  any  parallelogram  inscribed  in  a 
circle  is  a  rectangle. 

6.  What  is  the  locus,  of  all  the  vertices  of  right-angled  tri- 
angles erected  on  a  common  hypotenuse  ? 

7.  Prove  that  the  opposite  angles  of  any  inscribed  quad- 
rilateral are  supplementary. 

8.  Prove  that  any  equilateral  polygon  inscribed  in  a  circle  i? 
also  equiangular. 


II,  §  137]  MEASUREMENT  OF  ANGLES  115 

136.  Theorem  XIII.  An  angle  formed  by  two  chords 
intersecting  loithin  a  circle  is  measured  by  one  half  of 
the  sum  of  the  intercepted  arcs. 


Fig.  99 


Outline  of  proof.     Draw  AD. 

Then 

/AKB=AD-{-^^A. 

§  64 

But 

ZD  is  measured  by  ^  AB 

§  132 

and 

ZAis  measured  by  ^  CD ; 

§  132 

hence 

Z  AKB  is  measured  by  i  {AB  +  CD). 

Why? 

137.  Theorem  XIV.  An  angle  formed  by  a  tangent  and  a 
chord  drawn  through  the  point  of  tangency  is  measured  by  one 
half  of  the  intercepted  arc. 


C 


B. 


Outline  of  proof.     Draw  AE  II  CD. 

Then  AB  =  BE  §119 

and  ZABC=ZA.  §54. 

But  ZAis  measured  by  |  BE;  Why  ? 

hence  Z  ABC  is  measured  by  ^  AB. 


116 


THE  CIRCLE 


[II,  §  138 


138.  Theorem  XV.  An  angle  formed  by  two  secants,  or 
by  a  tayigent  and  a  secant,  or  by  two  tangents  that  meet  outside  a 
circle  is  measured  by  one  half  the  difference  of  the  intercepted  arcs. 


Fig.  101 

[The  proofs  are  left  to  the  student.     Draw  DE  parallel  to  BC  and 
note  how  the  angle  ADE,  which  is  equal  to  C,  is  measured.] 


EXERCISES 

1.  Prove  Theorem  XIII  by  drawing  a  line  through  A 
(Fig.  99)  parallel  to  BD. 

2.  If,  in  Fig.  99,  the  arc  BC  contains  130°  and  the  arc 
ABCD  contains  170°,  how  many  degrees  are  there  in  the 
angle  AKB  ?  Ans.  20°. 

3.  Two  angles  of  an  inscribed  triangle  are  70°  and  91°. 
Find  in  degrees  the  arcs  subtended  by  each  of  the  sides. 

4.  A  chord  that  divides  a  circumference  into  arcs  one  of 
which  contains  75°  is  met  at  one  extremity  by  a  tangent.  At 
what  (acute)  angle  does  the  meeting  take  place  ? 

5.  A  chord  is  met  at  one  extremity  by  a  tangent,  making 
with  it  an  angle  of  61°.  Into  what  arcs  does  the  chord  divide 
the  circumference  ? 

6.  If  a  tangent  is  drawn  at  the  vertex  of  an  inscribed  square, 
•how  many  degrees  are  there  in  the  angle  included  between  the 

tangent  and  a  side  of  the  square  ?     Answer  the  same  question 
for  an  inscribed  equilateral  triangle. 


II,  §  139] 


CONSTRUCTION   PROBLEMS 


117 


PART  IV.  CONSTRUCTION  PROBLEMS 

139.   Problem  1.     Through  a  given  point  to  draw  a 
tangent  to  a  circle. 

Case  1.      When  the  point  is  on  the  circumference. 

,A 


i?"lQ.  102 

Given  the  O  0  and  the  point  P  on  the  circumference. 
Required  to  draw  a  tangent  to  the  O  0  through  P. 
Construction.     Draw  the  radius  OP.     At  P  draw  AB  ±  OP. 
Then  AB  is  the  required  tangent.     Why  ? 

Case  2.      Wien  the  given  point  is  not  on  the  circumference. 

Outline  of  construction.     On  OP  as  diameter  draw  a  circle 
cutting  the  given  circle  at  A  and  B.  ^ 

Pass  a  line  through  the  points 
P,  A,  and  another  line  through  the 
points  P,  B. 

Either  of  these  lines  is  a  tangent 
such  as  desired.    Why  ?    In  answer- 
ing, note  that  Z  PAO  and  Z  PBO  ^^-^^^_    ^^. 
are  right  angles  and  apply  §  115.  fiq.  io3 


EXERCISES 

1.  Show  how  to  construct  a  tangent  to  a  given  circle  parallel 
to  a  given  line. 

2.  Show  how  to  draw  a  tangent  to  a  given  circle  perpen- 
dicular to  a  given  line. 


118  ,  THE  CIRCLE  [II,  §  140 

140.  Problem  2.  To  circumscribe  a  circle  about  a  given  tri- 
angle. (See  §  120.  The  student  should  carefully  state  the 
construction  and  the  proof.) 


C 


Fig.  104 

141.     Problem  3.     To  inscribe  a  circle  in  a  given  triangle. 
(See  §  124.     State  carefully  the  construction  and  its  proof.) 

C 


EXERCISES 

1.  Can  a  circle  always  be  drawn  tangent  to  three  lines,  no 
two  of  which  are  parallel  ?     If  so,  how  ? 

2.  Construct  an  equilateral  triangle  each  of  whose  sides 
equals  2  inches,  and  then  construct  its  inscribed  circle.  Prove 
that  in  this  case  the  center  of  the  circle  lies  at  the  inter- 
section of  the  three  altitudes  of  the  triangle.  Can  the 
same  statement  be  made  for  any  other  triangle? 

3.  In  Fig.  105,  would  the  lines  BI  and  AI  (extended)  ever 
pass  through  the  points  of  contact  M,  L  between  the  inscribed 
circle  and  the  sides  AC  and  BCj  respectively  ?  If  so,  when 
would  they,  and  why  ? 


II,  §  142] 


CONSTRUCTION   PROBLEMS 


119 


142.   Problem  4.     On  a  given  straight  line  to  construct  a  seg- 
ment  of  a  circle  that  shall  contain  a  given  angle. 

Given  line  AB  and  Z  x. 

Required  to  construct  on 
AB  a  segment  of  a  circle  that 
shall  contain  the  Z  x. 

Construction.    Construct  Z  ABC=  Z  x. 

Draw  ED  ±  AB  at  its  middle  point  D. 

Draw  FB  ±GBsitB  and  meeting  ED 
in  O. 

Draw  a  circle  with  0  as  center  and 
OB  as  radius. 

Then  AliB  is  the  segment  required. 

Proof.     Z  ABB  =  Z  ABC  =  Z  x.     Why  ? 


Fig.  106 


EXERCISES 

1.  On  a  line  2  inches  long  construct  a  segment  of  a.  circle 
that  shall  contain  an  angle  of  60°.     Do  the  same  for  30° ;  135°. 

2.  Construct  the  locus  of  the  vertices  of  all  triangles  hav- 
ing a  common  base  2  inches  long  and  a  common  vertex  angle 
of  30°. 

3.  In  Ex.  2,  p.  114,  what  can  be  said  of  the  locus  when  the 
angle  A  is  60°  instead  of  90°  ? 

4.  Construct  the  triangle  whose  base  is  2  inches,  whose  alti- 
tude is  3  inches,  and  whose  vertex  angle  is  30°. 

[Hint.  First,  draw  the  segment  of  a  circle  on  a  base  of  2  inches,  and 
such  that  30°  is  contained  in  it.  Now,  draw  a  parallel  to  the  same  base 
at  a  distance  of  3  inches  above  it.  Where  this  parallel  cuts  the  circle 
bounding  the  segment,  will  be  the  vertex  of  the  triangle  desired.  Why  ? 
The  desired  triangle  is  now  easily  drawn.] 

5.  How  many  solutions  will  there  be  to  any  problem  similar 
to  Ex.  4  ?  May  there  be  only  one  solution  ?  May  there  be  no 
solution  ?     Describe  all  possible  cases. 


120  THE  CIRCLE  [II,  §  142 

MISCELLANEOUS  EXERCISES.     CHAPTER  II 

1.  Prove  that  the  construction  described  in  §  5  for  drawing 
a  perpendicular  to  a  straight  line  at  a  point  within  it  is  correct. 

2.  Prove  that  the  constructions  of  §§  6  and  8  are  correct. 

3.  Through  a  given  point  in  a  circle  construct  the  chord 
that  is  bisected  by  that  point. 

4.  If  a  right  triangle  be  inscribed  in  a  circle,  show  that  its 
hypotenuse  will  be  a  diameter. 

5.  Prove  that  an  exterior  angle  of  an  inscribed  quadrilateral 
equals  the  opposite  interior  angle. 

6.  Prove  that  the  angle  between  two  tangents  to  a  circle  is 
double  the  angle  between  the  chord  joining  the  points  of  con- 
tact and  the  radius  to  a  point  of  contact. 

7.  Construct  a  circle  such  that  it  shall  pass  through  two 
given  points  and  shall  have  its  center  on  a  given  line. 

8.  If  four  points  A,  B,  C,  D  lie  on  a  circle,  and  if  AB  —  CD 
and  AD=BC,  show  that  the  lines  ^(7  and  BD  meet  at  the  center. 

9.  Prove  that  the  shortest  line  from  a  point  to  a  circum- 
ference is  along  the  radius  through  the  point,  extended  if 
necessary,  (a)  when  the  point  is  outside  the  circle ;  (&)  when 
the  point  is  inside  the  circle. 

10.  Prove  that  the  angle  between  two  tangents  to  a  circle  is 
the  supplement  of  the  angle  between  the  radii  drawn  to  the 
points  of  tangency. 

VI 1.   If  the  sides  of   an  inscribed  angle  are  parallel  to  the 
sides  of  a  central  angle,  how  do  their  arcs  compare  ? 

12.  Show  that  two  tangents  to  the  same  circle  are  parallel 
if  and  only  if  their  points  of  tangency  lie  at  opposite  ends  of 
the  same  diameter. 

13.  Show  that  a  circle  whose  diameter  is  one  side  of  any 
given  triangle  passes  through  the  feet  of  the  altitudes  drawn 
to  each  of  the  other  sides. 


II,  §  142]  MISCELLANEOUS  EXERCISES 


121 


14.  Following  the 
suggestions  of  the  ad- 
joining figures,  show 
how  to  construct  four 
common  tangents  to 
two  given  circles  if  the 
circles  do  not  intersect. 


Note.     The  shadows  thrown  by  a  round  body  illuminated  by  a  round 
source  of  light  (e.g.  earth  and  sun)  illustrate  this  exercise. 


122 


THE  CIRCLE 


[II,  §  142 


15.   How  many  common  tangents  can  be  drawn  to  two  given 
circles  in  each  of  the  cases  illustrated  in  the  following  figures  ? 

A 


what 
Find  approximately,  from  an 


16.  A  common  railroad  turn  is  made  by  an  arc  of  a  circle 
tangent  to  two  straight  portions  (Ex. 
5,  p.  101).     In  the  adjoining  figure,  if 
OG  =  OF,  and  OKI.  BF,  show  that : 

(1)  ZOGB=Z.FOB/2=ZCOB/4:, 

(2)  OJTis  parallel  to  BG. 

(3)  BK  =  BC/4., 
and  Z  0GB  =  Z  FOH. 

(4)  ZEBF=ZBOK, 
and  Z.EBC=ZBOF. 

17.  If  in  Ex.  16,  ZCOB=60°, 
is  the  length  of  the  chord  BC  ? 
accurate  figure,  or  from  the  table  of  chords,  p.  iii,  the  length 
of  the  chord  BF.     Prove  that  BF  =  FC  >  BC/2. 

Note.  In  practice  the  entire  curve  BC  can  be  (and  frequently  is)  laid 
off  by  means  of  a  surveying  instrument  located  at  B,  by  knowing  accurately 
the  lengths  of  a  number  of  such  chords  through  B,  and  their  directions. 

I      -* 

18.  Construct  a  circle  through  a  given  point 
tangent  to  a  given  line  at  a  given  point.  p 


19.  Construct  the  triangle  whose  base  is  2  inches,  whose  me- 
dian to  the  same  base  is  3  inches,  and  whose  vertical  angle  is  30°. 

[Hint.  The  vertex  must  lie  at  the  intersection  of  two  circles,  one 
bounding  the  segment  mentioned  in  the  Hint  to  Ex.  4,  p.  119,  and  the  other 
described  about  the  middle  point  of  the  base  with  a  radius  of  3  inches.] 


TI,  §  142 


MISCELLANEOUS  EXERC  ISES 


123 


20.  Show  that  a  quadrilateral  can  be  in- 
scribed in  a  circle  if  the  sum  of  one  pair  of 
its  opposite  angles  is  180°, 

21.  Show  that  a  kite  (see  Ex.  30,  p.  87)  that 
is  inscribed  in  a  circle  has  two  right  angles; 
hence  show  that  one  diagonal  is  a  diameter. 

22.  A  simple  turnout  off  of  a  straight  track  on  street  railways 
is  of  the  form  of  a  circular  arc, 
as  shown  in  the  figure.  The 
length  i2=0^=0C=  the  radius 
of  the  outer  rail  of  the  circular 
track,  is  called  the  radius  of  the 
turnout.  The  distance  L  =  BC 
from  the  point  of  the  switch  B 
to  the  crossing  C  is  called  the 
lead,  and  the  angle  TCS  between 
the  tangent  to  the  circular  track  at  G  and  the  straight  track,  is 
called  the  angle  of  crossing. 

Show  that  if  R  is  given  (as  it  usually  is),  the  angle  SCT 
of  crossing  (which  must  be  known  accurately  for  the  purpose 
of  ordering  the  proper  rail  cross)  can  be  found  from  the  figure 
by  measuring  the  angle  COA. 

23.  A  double  opposite  turnout  from  the 
end  of  a  straight  track  is  composed  of  two 
circular  portions,  the  radii  of  the  outer  rails 
being  given  lengths  r  and  R,  respectively, 
and  these  outer  rails  being  tangent  to  the 
straight  track  from  which  they  start.  Show 
that  the  angle  of  crossing  SCT  is  the  sum 
of  the  two  angles  COA  and  CO'B. 

24.  Let  D  be  any  point  on  the  side  AC  of  a  right  triangle 
ABC,  and  let  E  be  the  intersection  of  the  hypotenuse  AB 
with  a  circle  whose  diameter  is  AD.  Show  that  the  quadri- 
lateral DCBE  is  inscribed  in  the  circle  whose  diameter  is  DB, 


124 


THE  CIRCLE 


[H,  §  142 


25.  If  from  one  corner  ^  of  a  paral- 
lelogram ABCD  as  center,  a  circle  of 
radius  AB  is  drawn,  meeting  AB  pro- 
duced at  L,  and  the  sides  AC  and  BD 
at  M  and  N,  respectively,  show  that 
LM=  MN. 

26.  The  usual  form  of  carpenter's  level  contains  a  glass  tube 
filled  with  liquid,  except  for  a  small  bubble  of  air.  The  inner 
surface  of  the  tube  is  cut  in  the 
form  of  a  circular  arc  LHL', 
whose  center  is  at  0;  and  it 
is  set  into  a  wooden  or  metal 
frame  whose  top  edge  is  paral- 
lel to  the  tangent  at  H.  Show 
that  when  the  level  is  tilted 
through  the  angle  A,  the  bub- 
ble moves  through  an  arc  HN 
which  subtends  the  same  angle  A  at  the  center  0. 
level  be  more  sensitive  if  the  radius  R 
(=  OH)  is  large  or  small  ? 

27.  If  two  circles  intersect,  show  that 
the  obtuse  angle  between  the  two  tan- 
gents to  the  two  circles  at  either  point 
of  intersection  is  the  same  as  the  obtuse 
angle  between  the  two  radii  drawn  to 
that  point.  Hence  show  that  the  angle 
between  the  two  tangents  at  one  point  of  intersection  is  the 
same  as  that  between  the  two  tangents  at  the  other  point. 

28.  Prove  that  the  sum  of  two  opposite  sides  of  a  circum- 
scribed quadrilateral  equals  the  sum  of  the  other  two  sides. 

29.  Show  how  to  construct  the  common  tangents  to  the  pairs 
of  circles  represented  in  (1),  (2),  (3),  Ex.  15,  p.  122. 

30.  Construct  a  rectangle  having  given  a  side  and  the  diagonal 


Will  the 


II,  §  142] 


MISCELLANEOUS  EXERCISES 


125 


31.  Construct  an  isosceles  triangle  having  given  the  base 
and  the  length  of  the  altitude  from  one  end  of  the  base  to  one 
of  the  equal  sides. 

32.  Construct  a  circle  whose  center  lies  on  one  side  of  a 
given  triangle,  and  which  is  tangent  to  each  of  the  other  two 
sides. 

1.33.  Construct  a  triangle,  having  given  the  base  and  the 
altitudes  to  the  other  two  sides. 

34.  Construct  a  triangle, 'having  given  the  base,  the  altitude 
to  the  base,  and  a  second  altitude. 

35.  Construct  a  right  triangle, 
from  the  suggestion  contained  in 
the  figure,  when  the  hypotenuse 
AB  and  the  sum  AD  of  the  two 
sides  are  given. 

^6.  Show  that  the  bisectors  of  opposite  angles  of  a  parallelo- 
gram are  parallel  or  coincide.  Hence  show  that  the  only  kind 
of  parallelograms  in  which  a  circle  can  be  inscribed  is  a 
rhombus. 

37.  Show  that  the  figure  formed  by  two  tangents  PA  and 
PB  to  a  circle  O  and  the  radii  OA 

and  OB  forms  a  kite  which  is  in- 
scribed in  a  circle  0'  whose  diam- 
eter is  OP. 

Let  00'  meet  the  circle  0  at  (7 
and  D.  Show  that  Z  CO  A  is  meas- 
ured by  half  the  arc  AP ;  and  Z  CBA 
by  one  fourth  the  arc  AP. 

38.  In  the  figure  for  Ex.  37,  draw  the  radius  O'A.  Show 
that  any  change  in  Z  CDA  causes  four  times  as  much  change 
in  Z  PO'A 

39.  Show  what  becomes  of  the  figure  of  Ex.  37  if  the  angle 
COA  is  equal  to  90°. 


CHAPTER  III 

PROPORTION       SIMILARITY 

PART   I.     GENERAL  THEOREMS   ON  PROPORTION 

143.  Definitions.  An  equality  of  two  ratios  is  called  a 
proportion. 

Thus,  the  ratio  2/3  being  equal  to  the  ratio  4/6  gives  us  the 
proportion  2/3  =  4/6.  More  generally,  if  the  ratio  ajh  is  equal 
to  the  ratio  c/d,  then  the  equality  ajh  =  c/d  is  a  proportion. 

The  form  -  =  -  is  identical  with  a/&  =  c/d,  since  aJh  is  only  a  con- 
6      d 

yenient  way  of  printing  a  fraction. 

Besides  writing  a  proportion  in  the  form   aJh  =  c/rf,   either  of  the 

following  forms  may  also  be  used,  a  :  &  =  c  :  <Z,  or  a  :  6  : :  c  :  d     In  any 

case,  the  proportion  is  read  ^^  a  is  to  b  as  c  is  to  (?." 

In  the  proportion  a:b  =  c:  d,  the  numbers  a,  b,  c,  and  d  are 
the  terms.  The  first  and  fourth  of  these  (a  and  d)  are  called 
the  extremes,  while  the  second  and  third  (b  and  c)  are  called  the 
means.  Again,  the  first  and  third  (a  and  c)  taken  together 
are  called  the  antecedents,  while  the  second  and  fourth  taken 
together  are  called  the  consequents. 

A  series  of  equal  ratios  in  the  form  a/b  =  c/d  =  e/f=  •••is 
called  a  continued  proportion.     Thus,  2/3  =  4/6  =  8/12  =  •••. 

EXERCISES 

1.  Using  the  language  of  proportion,  read  the  equation 
f  =  \^.  What  are  the  extremes  here,  what  the  means ;  which 
the  antecedents  and  which  the  consequents  ? 

126 


Ill,  §  144]  GENERAL  THEOREMS  127 

2.  Form  and  read  off  such  proportions  as  you  can  make  out 
of  the  following  four  quantities :  2  in.,  8  in.,  4  in.,  16  in.  Do 
the  same  for  1  in.,  3  in.,  1  ft.,  1  yd. 

3.  What  number  bears  the  same  ratio  to  2  as  8  does  to  3  ? 

[Hint.  In  this  and  in  all  similar  questions,  let  x  represent  the  desired 
number  and  form  an  equation.  Thus,  we  here  have  a;/2  =  8/3.  Now 
solve  this  equation  for  x.] 

4.  Find  two  numbers  such  that  their  ratio  is  as  3:5  and 
whose  sum  is  4. 

[Hint.  Let  x  and  y  represent  the  unknown  numbers  and  form  equations.] 

144.  General  Theorems  on  Proportion. 

Theorem  A.  In  any  jwoportioRf  the  product  of  the  extremes 
is  equal  to  the  product  of  the  means. 

Given  a/b  =  c/d,  to  prove  that  ad  =  be. 

Proof.  Clear  the  given  equation  of  fractions  by  multiplying 
both  its  members  by  bd.  The  result  is  the  desired  equation 
ad  =  be. 

Corollary  1.  If  the  two  antecedents  of  a  iiroportion  are  equal, 
the  consequents  are  also  equal. 

Theorem  B.  If  the  product  of  two  num.bers  is  equal  to  the 
product  of  two  other  numbers,  either  pair  may  be  made  the  means 
of  a  proportion  in  which  the  other  two  are  taken  as  the  extremes. 

Given  ad  =  be,  to  prove  that  a/b  =  c/d. 

Proof.  Divide  both  members  of  the  given  equation  by  bd. 
In  order  to  prove  that  b/a  =  d/c,  write  the  given  equation  in 
the  form    be  =  ad  and  then  divide  both  members  by  ac. 

Theorem  C.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  by  inversion ;  that  is,  the  second  term  is  to  the  first 
as  the  fourth  is  to  the  third. 

Given  a/b  =  c/d,  to  prove  that  b/a  =  d/c. 

Proof,  ad  =  be  by  Theorem  A ;  hence  b/a  =  d/c  by  Theo 
rem  B. 


128  PROPORTION      SIMILARITY  [III,  §  144 

Theorem  D.  If  four  quantities  are  in  proportion^  they  are  in 
proportion  by  alternation ;  that  is,  the  first  term  is  to  the  third 
as  the  second  is  to  the  fourth. 

[The  proof  is  left  to  the  student.    First  use  Theorem  A.] 

Theorem  E.  If  four  quantities  are  in  proportion,  they  are  vn 
proportion  by  composition;  tliat  is,  the  sum  of  the  first  two 
terms  is  to  the  second  term  as  the  sum  of  the  last  two  terms  is 
to  the  last  term. 

Given  a/b  =  c/d,  to  prove  that  (a  +  b)/b  =  (c  +  d)/d. 

Proof.     Adding  1  to  each  side  of  the  given  equation,  we  have 

«  +  l  =  ^  +  l,or^  +  *  =  «-+^.  Why? 

b  d  b  d 

[Show  in  a  similar  way  that  (a  +  b)/a  =  (c  +  d)/c.'] 

Theorem  F.  If  four  quantities  are  in  proportion,  they  a.re  in 
proportion  by  division ;  that  is,  the  difference  between  the  first 
two  terms  is  to  the  second  term  as  the  difference  between  the 
last  two  terms  is  to  the  last  term. 

Given  a/b  =  c/d,  to  prove  that  (a  —  b)/b  =  (c  —  d)/d. 

Outline  of  proof.     Make  use  of  the  fact  that  a/b  —  1  =  c/d  —  1. 

[Show  in  a  similar  way  that  («  --  h)/a  =  (c  ■-  d)/c.'] 

Theorem  G.  If  four  quantities  are  in  proportion,  they  are 
in  proportion  by  composition  and  division ;  that  is,  the  sum  of 
the  first  two  terms  is  to  their  difference  as  the  sum  of  the  last 
two  terms  is  to  their  difference. 

Given  a/b  =  c/d,  to  prove  that  {a-\-b) / (a  —  b)  =  {c-\-d) / {c—d). 
Proof.     We  have 

a±b^c_±d    an^a-6^c-d.  Th.  E,  F 

b  d    '  b  d 

Therefore     {a  +  b)/{a-b)=^{c  +  d)/{c-d).  Ax.   4 


Ill,  §  144]  GENERAL  THEOREMS  129 

Theorem  H.  In  a  series  of  equal  ratios  the  sum  of  the  ante- 
cedents is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its 
consequent. 

Given  a/6  =  c/d  =  e//=  •••,  to  prove  that 

b-\-d-{-f+'"~b~d  ""/""*' 
Proof.     Let  k  be  the  value  of  any  one  of  the  equal  ratios 


so  that 

Then 
hence 

or, 


ace 
b'  d'  f 


lc  =  ^  =  '-  =  ±^ 


h      d     f 

a  =  kb,  c  =  kd,  e  =  kf,  — ;  Why  ? 

b-\-d-^f-h"'         6  +  d  +  /+... 

a4-c  +  eH —  _  ^  _  2.  _  1  —  ... 
b  +  d+f+'"~b~'d~f'~"'' 

EXERCISES 

1.  Given  the  proportion  1/2  =  3/6.  Write  down  the  various 
proportions  which  come  from  this  by  (1)  inversion,  (2)  alter- 
nation, (3)  composition,  (4)  division,  (5)  composition  and  divi- 
sion.    Note  that  each  one  thus  obtained  is  a  true  proportion. 

2.  If  a/b  =  c/d,  prove  that  ma/mb  =  c/d,  where  m  represents 
any  number. 

3.  If  a/b  =  c/d,  prove  that  2  a/3  6  =  2  c/3  d ;  also  that 
(a-|-3  6)/6  =  (c-f-3d)/d 

4.  If  a/6  =  b/c,  prove  that  6^^  =  ac ;  also  that  a/c  =  a^/b^. 

5.  If  b^  =  ac,  prove  that  a/6  =  b/c. 

6.  If  a/6  =  b/c,  prove  that  c  =  6Ya. 


130  PROPORTION      SIMILARITY  [III,  §  145 

PART   II.     PROPORTIONAL   LINE-SEGMENTS 

145.   Theorem  I.    A  line  parallel  to  the  base  of  a  triangle 
divides  the  other  sides  proportionally. 


Fig.  107 

Given  the  A  ABC  and  the  line  DE  II  the  base  AB. 

To  prove  that  CD /DA  =  CE/EB. 

Proof,     (a)    When  CD  and  DA  are  commensurable.       (§  128) 

In  this  case  a  unit  length  may  be  found  which  is  contained 
an  exact  number  of  times  in  both  CD  and  DA.  Let  m  be 
such  a  unit  and  let  us  suppose  that  this  unit  is  contained  r 
times  in  CD  and  s  times  in  DA. 

Then, 

CD  ^r 
DA     s' 


(1) 


Now  divide  CD  into  its  r  divisions,  each  of  length  m,  and 
through  the  points  of  division  draw  lines  parallel  to  AB. 
Do  the  same  with  DA. 

These  parallels  will  cut  CE  into  r  equal  parts  and  EB  into 
s  equal  parts.     (§§  91-92.)     Therefore 

(2)  ^=r. 

^  ^  EB     s 

Comparing  (1)  and  (2)  we  obtain,  as  was  to  be  proved, 

CD^CE 
DA     EB' 


Ill,  §  146]     PROPORTIONAL  LINE-SEGMENTS  131 

Note.  As  in  §  130,  this  essentially  completes  the  proof; 
and  the  theorem  holds  also  in  the  case  CD  and  DA  are  incom- 
mensurable, by  the  postulate  of  §  130. 

The  following  argument  establishes  the  truth  of  that  postu- 
late in  the  case  of  the  present  theorem ;  it  may  be  omitted  at 
the  discretion  of  the  teacher.  q 

Proof.     (6)  When  CD  and  DA  are   incom-  j\  _ 

mensurable.  I     \ 

In  this  case  if  we  take  any  unit  lengfh  m  I         \ 

which  is  contained  an  exact  number  of  times  /  \ 

in  CD  and  apply   it  to  DA,  there  will  remain      q/ ^^ 

after  the  last  point  of  division  a  certain  length        /  \ 

AF  which  will  be  less  than  m.  ^r  -\i 

But,  whatever  the  choice  of  m,  we  shall  have,  _,       .„„ 

by  case  (a),  CD/DF=  CE/EG.  ^^' 

Now,  as  m  is  taken  smaller  and  smaller,  this  equation  holds  true  at 
every  step.  At  the  same  time  the  individual  members  of  the  same  equa- 
tion are  changing,  but  only  so  that  DF  comes  closer  to  DA,  while  EC 
comes  closer  to  EB.  By  taking  m  sufficiently  small  we  can  thus  bring 
CD / DF  and  CE/EG  as  near  as  we  please  to  the  respective  values 
CD/ DA  and  CE/EB.  These  last  ratios,  since  they  differ  by  as  little  as 
we  please  from  the  preceding  equal  ratios,  must  differ  from  each  other  by 
as  little  as  we  please.  This  is  the  same  as  saying  that  they  are  actually 
equal ;  for,  if  unequal,  their  difference  could  not  by  any  method  of 
reasoning  be  made  as  small  as  we  please,  since  they  are  fixed  quantities. 

Thus  we  must  have,  as  was  to  be  proved,  CD /DA  =  CE/EB. 

146.  Corollary  1.  If  a  line  is  drawn  parallel  to  the  hose  of 
a  triangle,  either  side  is  to  one  of  its  segments  as  the  other  side  is 
to  its  corresponding  segment. 

Given  the  A  ABC  and  DE  II  AB.     (See  Fig.  107.) 
To  prove  (1)  that  CA/CD  =  CB/CE 
and  (2)  that  CA/DA  =  CB/EB. 

[Outline  of  proof.  Having  shown  in  Theorem  I  that  CD/ DA  =  CE/EB^ 
we  obtain  by  composition  (E,  §  144)  {CD-\-DA)/DA  =  {CE-\-EB)/EB, 
from  which  (1)  follows.  To  prove  (2),  taking  the  proportion  CD/ DA 
=  CE/EB  by  inversion  (C,  §  144)  we  have  DA/ CD  =  EB/CE.  Now 
take  this  proportion  by  composition.] 


132 


PROPORTION    SIMILARITY 


[in,  §  146 


EXERCISES 


1.    In    the    adjacent    figure    DE 


CD  =  2  in.,  CA  =  2>  in.,  and  CE 


IS 

3* 


parallel 
in., 


Ans.  If  in. 


what  is  the  length  of  EB? 

2.  If,  in  the  figure  of  Ex.  1,  we  have 
CA  =  %  ft.,  05  =  12  ft.  3  in.,  and  CE  =  ^ 
ft.,  what  is  the  length  of  DA  ? 

3.  A  thread  DE  is  stretched  in  any  direction  across  a  tri- 
angle ABC  cutting  two  of  its  sides  in  the  C 
points  D,  E.      The  thread  is  now  moved                 y^\£. 

up  and  down  over  the  surface  of  the  tri- 
angle, but  always  in  such  a  way  as  to  be 
parallel    to    its   original   direction.      How    A  B 

does   the  ratio  CD/CE  change  during  the  motion  ?     Why  ? 

4.  Prove  that  any  line  drawn  par- 
allel to  the  base  of  a  trapezoid  cuts 
the  other  two  sides  proportionally. 

5.  If   a   common   exterior    tangent 
to  two  circles  (see  figures  for  Ex.  14, 
p.  121)  meets  the  line  of  centers  00' 
in  a  point  G,  show  that  the  lengths  along  this  tangent  from 
G  to  the  two  circles  are  in  the  ratio  OG/O'G, 

6.  A  channel  for  water  has 
a  cross-section  of  the  form 
of  a  trapezoid  ABCD.  If 
BC  and  AD  are  horizontal, 
show  that  the  wetted  por- 
tions of  the  side  walls  BE  B  C 
and  CF  are  proportional  to  the  dry  portions  AE  and  DF. 

7.  What  other  proportions  can  you  write  down  for  Ex.  6  ? 

8.  In  Ex.  6,  if  AB  =  15  ft.,  CD  =  12  ft.,  and  the  vertical 
height  H  from  BC  to  AD  is  10  ft.,  find  the  lengths  of  the  dry 
portions  of  the  side  walls  when  the  water  is  4  ft.  deep. 


Ill,  §  148]     PROPORTIONAL  LINE-SEGMENTS  133 

147.  Theorem  II.  {Converse  of  Theorem  I.)  If  a 
line  divides  two  sides  of  a  triangle  proportionally,  it 
is  parallel  to  the  third  side. 

c 


Fig.  109 

Given  the  A  ABC  and  the  line  DE  dividing  the  sides  pro- 
portionally ;  that  is,  such  that  CD/DA=CE/EB. 

To  prove  that  BE  is  parallel  to  AB. 

Proof.  Suppose  that  DE  is  not  II  AB.  Then  draw  DE'  II  AB. 
We  should  now  have 

(1)  CA/CD  =  CB/CE'.  §  146 

Also,  since  by  hypothesis 

CD/DA^CE/EB, 
it  follows  that 

(2)  CA/ CD  =  CB/CE.  Th.  E,  §  144 

From  equations  (1)  and  (2)  we  should  have 

CE'  =  CE ;  Th.  A,  Cor.  1,  §  144 

whence  DE'  would  coincide  with  DE.  Post.  1 

This  is  impossible  unless  DE  II  AB.  Why  ? 

Therefore,  DE  II  AB. 

148.  Corollary  1.  If  a  line  cuts  two  sides  of  a  triangle  in  such 
a  ivay  that  either  side  is  to  one  of  its  segments  as  the  other  side  is 
to  its  corresponding  segment,  then  the  line  is  parallel  to  the  third 
side 

[Hint.  Show  that  under  the  given  conditions  the  hypothesis  of  §  147 
is  satisfied.    Compare  with  §  89  and  §  146.] 


134  PROPORTION      SIMILARITY         [III,  §  149 

149.  Theorem  III.  The  bisector  of  an  angle  of  a  triangle 
divides  the  opposite  side  into  segments  ivhich  are  proportional  to 
the  sides  of  the  angle. 

Given   the  A  ABC  and   the   bi- 
sector CD  of  Z  C 

To  prove  that  AD/DB  =  AC/BG. 

Proof.    Draw  BE  il  CD  and  meet- 
ing AC  prolonged  at  E. 

Then  AD/DB  =  AC/ CE.  Why? 

Moreover,  Zx  =  Zyj     Z  x=  Zw,     Zy  =  Zz.  Why  ? 

Therefore,  Zz  =  Zw;  whence  CE  =  BC ;  Why ? 

hence  AD/DB  =  AC/BC.  Ax.  9 

EXERCISES 

1.  In  the  triangle  ABC  of  Fig.  110,  CA  =  11  in.,  CB=S  in., 
and  AB  =  14  in.  What  are  the  lengths  of  the  segments  AD 
and  DB  into  which  the  bisector  CD  of  the  angle  C  divides 
the  base  AB  ?  Ans.  AD  =  8^%  in.,  DB  =  5{i  in. 

I  2.  Prove  by  means  of  §  147  that  the  line  joining  the  middle 
points  of  two  sides  of  a  triangle  is  parallel  to  the  third  side 
and  equal  to  half  of  it.     (Compare  with  §  89.) 

!  3.    State  and  prove  the  converse  of  Theorem  III. 

[Hint.  Prolong  AC  to  E,  making  CE  =  BC.  Draw  EB.  (Fig.llO.) 
Prove  AC/AD=  CE/DB,  and  hence  CD  WEB.  Then  Zx  =  Zw  and 
Zy  =  Zz.     But  Zw=Zz   (since  CB  =  CE) .     Therefore  Zx=Zy.] 

4.  An  ordinary  closed  rubber  band  is  stretched  out  and  held 
against  a  board  by  means  of  two 
pins  A  and  B.  One  of  the  halves 
of  the  band  is  stretched  by  means 
of  a  pencil  point  into  the  form  of 
an  elastic  triangle  ABC,  with  the 
pencil  point  at  C.  How  must  the  pencil  point  move  in  order 
that  the  ratio  CA/CB  shall  remain  constant  ? 


Ill,  §150]      PROPORTIONAL  LINE-SEGMENTS  135 

150.   Theorem  IV.    If  a  series  of  parallels  be  cut  by  two 
lines,  the  corresponding  segments  are  proportional. 

— w  \^  X 

•  c/        \f     \j 


\E 


\ 


Fig.  Ill 


Given  the  parallels  AH^  BI,  CJ,  etc.,  cut  by  the  lines  AD  and 
HK. 

To  prove  that  AB/HI=  BG/IJ=  CD/JK. 

Proof.     Draw  AE  II  HK. 

Now  AC/AF  =  AB/AO  =  BC/GF  Why  ? 

and  AC/AF=CD/FE.  Why? 

Therefore         AB/AG  =  BC/GF  =  CD/FE.  Why  ? 

But        AG  =  HI,  GF  =  IJ,  FE  =  J/iT;  Why  ? 

hence  ^5/iJ/  =  BC/IJ  =  CD/ JiT.  Ax.  9 

Note.  All  work  on  proportional  segments  is  based  on  §§  145, 
150.     See  §§  149,  151,  153,  etc.     Compare  also  §§  91-94. 

EXERCISES 

1.  Prove  Theorem  IV  when  the  two  lines  AD,  HK  (Fig.  Ill) 
intersect  at  some  point  between  the  parallels  AH,  DK. 

2.  Prove  the  fact  stated  in  Cor.  1,  §  92  by  means  of  §  150. 

3.  If  the  three  segments  cut  out  of  one  transversal  by  four 
parallel  lines  are  2  in.,  3  in.,  4  in.,  respectively,  what  can  be 
inferred  concerning  the  perpendicular  distances  between  the 
parallels  ?  If  the  distance  between  the  first  pair  is  1  in.,  what 
are  the  distances  between  the  others  ? 


136  PROPORTION      SIMILARITY  [III,  §  151 

151.  Problem  1.     To  divide  a  given  line  into  parts 
proportional  to  any  number  of  given  Ihies. 


1       * 

A^.^ \ '. ^ .B 

^  ^-^    \        \           \ 

,           b 

d 

D" 

Fig.  112  ■ 

Given  the  line  AB. 

Required  to  divide  AB  into  parts  which  shall  be  proportional 
to  the  given  lengths  a,  b,  c,  d. 

Construction.     Draw  AC  making  any  angle  with  AB. 

Measure  off  AG  =  a,  GF=  b,  FE  =  c,En  =  d. 

Join  the  last  point  D  thus  determined  with  B,  and  through 
E,  F,  and  G  draw  parallels  to  BD  cutting  AB  in  H,  I,  and  J. 

Then  the  segments  AJ,  JI,  IH,  HB,  divide  AB  as  required. 

Proof.  AJ/AG  =  JI/GF  =  IH/FE  =  HB/ED ;  §  150 
hence  AJ/a  =  Jl/b  =  IH/c  =  HB/d. 

Note.  This  problem  includes  as  a  special  case,  the  division 
of  a  given  line  into  any  number  of  equal  parts  ;  for  if  we  take 
a  =  5  =  c  =  c?,  we  have  AG  =GF  =  FE=^  ED. 

EXERCISES 

1.  Divide  a  line  10  in.  long  into  8  equal  parts  by  Note,  §  151. 
Show  how  to  do  the  same  exercise  by  §  8. 

2.  Divide  a  line  10  in.  long  into  7  equal  parts  by  §  151. 
Can  this  exercise  be  done  by  §  8  ? 

3.  State  when  §  8  can  be  used  in  place  of  Note,  §  151  to 
divide  a  given  line  into  a  number  of  equal  parts. 


Ill,  §  152]     PROPORTIONAL   LINE-SEGMENTS  137 

4.  How  could  the  page  of  a  notebook  be  ruled  (accurately) 
into  three  equal  columns  ? 

5.  Divide  a  given  line  into  parts  which  are  to  each  other  as 
1:2:3. 

6.  Draw  a  line  AB  and  divide  it  into  fifths ;  thereby  find 
the  point  on  AB  one  fifth  the  distance  from  A  to  B.  Find  a 
pomt  2/5  of  AB  from  A. 

y  7.  Given  an  angle  ABC  and  a  point  P  any- 
where within  it.  To  construct  a  line  through 
P  such  that  the  portion  of  the  line  lying  within 
the  angle  shall  be  bisected  by  P. 

[Hint.    Through  P  draw  PD  II  BC.     Then  lay  off 
DE=DB.] 

'^  8.   Through  a  point  P  without  an  angle  ABG  to  draw  PE  so 
that  PJ9  =  Z>JS7. 

B 


9.  State  and  prove  a  theorem  concerning  the  segments  into 
which  the  non-parallel  sides  of  a  trapezoid  are  divided  by  a 
series  of  parallels  to  the  bases. 

10.  Layers  of  rock  in  the  earth  are  nearly  parallel,  but  they 
are  not  usually  level  (horizontal).  Show  that  the  ratios  of  the 
thicknesses  of  layers  of  different  kinds  of  rock  can  be  found 
by  measuring  the  thicknesses  of  the  cuts  made  by  a  vertical 
mine-shaft,  or  a  deep  well. 

152.  Definitions.  In  the  proportion  a/&  =  c/r7,  d  is  called 
the  fourth  proportional  to  a,  6,  and  c.  Thus,  6  is  the  fourth  pro- 
portional to  1,  2,  and  3.  If  6  =  c,  so  that  a/h  =  h/d,  d  is  called 
the  thiid  proportional  to  a  and  b. 


138  PROPORTION      SIMILARITY  [III,  §  153 

153.  Problem  2.     To  find  the  fourth  proportional  to 
three  given  lines. 


a 

A         a         C        b     B 

,         b       , 

-,  "^\^^.  \       \ 

c 

^^F 


Fig.  113 


Given  the  three  lines  a,  b,  and  c. 

Required  to  construct  their  fourth  proportional ;  that  is,  to 
find  a  line  x  such  that  we  shall  have  a/b  =  c/x. 

Construction.  Draw  a  line  AB  of  indefinite  extent  and  from 
A  lay  off  ^(7  equal  to  a  and  CB  equal  to  b. 

Draw  a  second  line  AF  of  indefinite  extent,  making  any  con- 
venient angle  at  A,  and  along  AF  lay  off  ^6r  equal  to  c. 

Join  CG.     Draw  BH II  CG  and  cutting  AF  at  //. 

Then  GH  is  the  required  line  x. 

[The  proof  is  left  to  the  student.     Use  §  145.] 

EXERCISES 

1.  Find  a  fourth  proportional  to  three  lines  2  in.,  4  in.,  and 
5  in.  long,  respectively.  Find  a  fourth  proportional  to  three 
lines  2  in.,  5  in.,  and  4  in.  long,  respectively. 

2.  Show  how  to  construct  a  third  proportional  to  two  given 
lines. 

3.  To  enlarge  a  line  segment  I  in  the  ratio  2/3  is  to  find  a 
new  line  segment  x  such  that  2/3  =  l/x.  Show  how  to  do  this 
geometrically. 

4.  Draw  a  triangle ;  then  enlarge  its  sides  in  the  ratio  o/5, 
using  geometric  constructions  only.     See  §  21,  p.  19. 


Ill,  §  154]  SIMILAR  TRIANGLES  139 

PART   HI.      SIMILAR   TRIANGLES   AND  POLYGONS 

154.  Definitions.  That  a  figure  may  be  enlarged  or  reduced 
in  size  without  essentially  changing  its  appearance  otherwise 
is  a  part  of  common  knowledge.  Thus  we  drew  such  enlarged 
and  reduced  figures  in  §  21.  Throughout  this  book  rather 
small  figures  are  printed;  these  the  student  almost  always 
enlarges  in  drawing  on  paper  or  at  the  board. 

In  general,  nearly  all  drawings  represent  objects  of  a  size 
different  from  that  of  the  drawing;  thus  house  plans  are 
usually  drawn  (§  21)  on  the  scale  of  -^  inch  to  one  foot. 

Enlargements,  or  reductions,  are  made  by  increasing,  or  de- 
creasing, all  lengths  in  a  figure  in  the  same  ratio  throughout, 
so  that  all  lengths  in  the  new  drawing  are  proportional  to  the 
corresponding  lengths  in  the  original  figure. 

Two  triangles  which  have  their  corresponding  sides  propor- 
tional are  called  similar.  Fig.  114  represents  two  similar 
triangles,  each  side  of  the  one  being  twice  that  of  the  other. 


A 


Fig.  114 


The  fact  that  two  triangles  ABC  and  A'B'C  are  similar  is 
expressed  by  the  relation  ABC  ^  A'B'C 

In  general,  if  one  figure  is  obtained  from  another  by  enlarge- 
ment or  reduction,  that  is,  if  all  lengths  that  can  be  drawn  in 
one  are  proportional  to  the  corresponding  lengths  in  the  other, 
the  two  figures  are  said  to  be  similar. 

Two  triangles  are  said  to  be  mutually  equiangular  when  their 
corresponding  angles  are  equal. 


lid  PROPORTION    Similarity        [iii,  §  155 

1S5.  Theoreiti  V.    If  two  triangles  are  mutually  equiangular, 
they  are  similar. 


a'^ ^b' 


Fig.  115 


Given  the  mutually  equiangular  A  ABC,  A'B'C 

To  prove  that      A  ABC  ~  A  A'B'C. 

Proof.  On  CA  lay  off  CD  =  C'A'  and  on  CB  lay  off  CE 
=  CB'.     Draw  DE.     Then,  in  the  A  DEC  and  A'B'C  we 

have                              CD  =  CA',  CE  =  CB',  Const, 

and                               ZC  =  ZC.  Given 

Therefore           A  DEC^  A  A'B'C.  Why? 

Whence  Z  Oi)J5;  =  Z  CM'5'. 

But                            ZA=Z  C'A'B'.  Given 

Hence                 Z  (7i)jE;  =  Z  A.  Why  ? 

It  follows  that          DE  II  ^5 ;  Why  ? 

,            1                       'CZ>      CE         CA'     CB'  „  .  .^ 

hence  also,                    —  =  —  or  -^  =  -^-  §  146 

CA'     A'B' 
In  like  manner  it  can  be  shown  that 


CA       AB 

Therefore  AABC^AA'B'C.  §154 

156.    Corollary  1.     Two  triangles  are  similar  if  two  angles  of 
the  one  are  equal  respectively  to  two  angles  of  the  other. 

^7157.   Corollary  2.     Two  right  triangles  are  similar  if  an  acute 
angle  of  the  one  is  equal  to  an  acute  angle  of  the  other. 


Ill,  §  158] 


SIMILAR  TRIANGLES 


141 


158.  Theorem  VI.     If  two  triangles  are  similar,  they 
are  mutually  equiangular,     {Converse  of  Theorem  F.) 


Fig.  116 
Given  the  two  similar  A  ABC  and  A'B'C ;  that  is,  two  A 
such  that -^  =  -^=-:^. 

C'A'    c'b:    a'b' 

To  prove  that  A  ABC  and  ABIC  are  mutually  equiangular. 
Proof.     On  CA  lay  off  CD=  C'A',  and  on  CB  lay  off  CE 
=  CB'.     Join  D  and  E. 

Then  CA/CD=CB/CE. 

Therefore,  DEW  AB\ 

whence,  Z  CDE=  Z  .4  and  Z  Z)^0  =  Z  5. 


Given 

§  147 
Why? 
It  follows  that  the  A  CDE  and  CAB  are  mutually  equi- 


angular ; 
whence  also 
therefore 

But 
whence 
and  hence 

Therefore 


Why? 

§155 

§154 

Given 

Why? 


A  CDE  --  A  CAB ; 
AB/DE=  CA/CD  =  CA/C'A' 
CA/C'A'  =  AB/A'B'', 
AB/DE  =  AB/A'B', 

DE=A'B'.  Th.  A,  Cor.  1,  §  144 

A  CDE  ^  A  C'A'B'.  Why  ? 

All  three  of  the  A  CAB,  CDE,  and  C'A'B'  are  therefore 
mutually  equiangular ;  hence,  in  particular,  the  A  ABC  and 
ABIC  are  mutually  equiangular. 


142       .  PROPORTION     SIMILARITY  [III,  §  158 

EXERCISES 

1.  Show  that  all  equilateral  triangles  are  similar. 

2.  Show  that  all  isosceles  triangles  that  have  the  same  ver- 
tex angle  are  similar. 

3.  Prove  that  if  each  of  two  triangles  are  similar  to  a  third, 
they  are  similar  to  each  other. 

4.  How  high  is  a  house  whose  shadow  is  144  ft.  long  when 
that  of  a  gate  post  5  ft.  high  is  12  ft.  long  ?  Ans.  60  ft. 

5.  If  a  triangle  be  stretched  out  so  that  each  side  becomes 
one  half  longer  than  at  the  beginning,  will  the  size  of  the 
angles  be  changed  ?     Why  ? 

6.  In  order  to  determine  the  length  AB  of  a  lake,  set  two 
stakes  at  convenient  points,  D  and  E,  such  that  AD  is  perpen- 
dicular to  AB,  while  DE  is 

perpendicular  to  AD.  Now, 
from  stake  E  sight  the  point 
B  and  take  note  of  the  point 
C  where  the  line  of  sight 
crosses  AD.  Show  how  to 
find  AB  by  measuring   ED,   DC,   and    CA. 

7.  Show  that  if  each  of  the  sides  of  a  triangle  is  reduced  or 
enlarged  in  the  same  scale  (§  21),  the  angles  are  unchanged. 
Thus  it  is  said  that  enlargement  does  not  distort  a  figure. 

8.  A  triangular  tin  plate  expands  in  the  same  ratio  along  all 
straight  lines  when  heated.     Show  that  its  shape  is  not  changed. 

9.  Prove  that  two  triangles  that  have  their  sides  parallel 
or  perpendicular  each  to  each  are  similar. 

10.  Prove  that  the  diagonals  of  any  trapezoid  divide  each 
other  proportionally. 

11.  Let  P  and  Q  be  any  two  points  on  the  sides  of  an  angle 
AOB.  Show  that  the  triangles  formed  by  perpendiculars  from 
P  and  Q  to  the  sides  opposite  them,  respectively,  are  similar. 


Ill,  §  159] 


SIMILAR  TRIANGLES 


143 


159.  Theorem  VII.  Tivo  triangles  are  similar  if  an 
angle  of  the  one  equals  an  angle  of  the  other  and  the  in- 
cluding sides  are  proportional. 

Given  the  A  ABC, 
A'B'C,  such  that  Z  C 
=  ZC  and  ■  CA/C'A' 
=  CB/C'B'. 

To  prove  that  A  ABC 
^  A  A'B'C. 

Proof     On    CA  lay   off     A'' ^B 

CD  =  CA\  and  on  CB  lay  Fig.  117 

off    CE  =  CB'     Join  D  and  E. 

Then  A  CDE  ^  A  CA'B'. 

Also  we  have  CA/CD^GB/CE; 

and  hence  DE  II  AB. 

Therefore     Z  CDE  =  Z.  A,  and  Z  DEC  =  Z  J5 ; 
whence    A  yl5(7  and  Z>^(7  are  mutually  equiangular; 
and  A  ABC  and   A'B'C  are    mutually  equiangular. 

Therefore  A  ^5(7  ~  A'B'C. 


§35 

Given 
§147 

Why? 

Why? 

Why? 
§155 


EXERCISES 

1.  Compare  the  various  conditions  under  which  two  triangles 
are  similar  with  those  under  which  two  triangles  are  congruent. 

2.  If  the  angle  at  the  vertex  of  a  triangle  is  held  fixed 
while  the  sides  which  include  the  same  angle  are  stretched  out 
by  half  their  length,  how  do  the  angles  in  the  resulting  tri- 
angle compare  with  those  in  the  original  triangle?     Why  ? 

3.  Two  rods  AB  and  CD  are  hinged  together  at  a  point  0, 
the  hinge  being  placed  on  each  rod  one  o 
third  the  length  of  the  rod  from  one  end.  ,^^ — .0 
Show  that  AC  =  BD/2,  whatever  the  a 
Z  DOB  may  be.  Show  also  that  the 
ratio  AD/CBdoes  not  change  when  Z  Z>OB changes.  liAB—  CD. 


144 


PROPORTION      SIMILARITY 


[in,  §  160 


160.  Definition.  If  in  the  proportion  a/h  =  c/d  we  have 
0  =  6,  then  h  (or  c)  is  said  to  be  the  mam  proportional  between 
a  and  d.  Thus,  2  is  the  mean  proportional  between  1  and  4, 
since  we  have  1/2  =  2/4. 

161.  Theorem  VIII.  If,  in  any  right  triangle,  a  perpendic- 
ular is  drawn  from  the  vertex  of  the  right  angle  to  the  hypote- 
nuse, the  two  right  triangles  thus  formed  are  similar  to  each 
other  and  to  the  given  triangle. 

Given  the  right-angled 
A  ABO  having  its  right  angle 
at  C.  Given  also  the  perpen- 
dicular CD  drawn  from  C  to 
the  hypotenuse  AB. 

To  prove  that  A  ADC,  CDB, 
and  ABC  are  similar. 

Proof.     These  A  are  mutually  equiangular ; 
therefore  they  are  similar. 


Fig.  118. 


Why? 
Why? 


162.  Corollary  1.  In  any  right  triangle  the  perpendicular 
from  the  vertex  of  the  right  angle  to  the  hypotenuse  is  the  mean 
proportional  between  the  segments  of  the  hypotenuse. 

Proof.  The  A  ADC,  CDB  being  similar,  their  corresponding 
sides  are  proportional.  Upon  comparing  the  sides  AD  and  CD, 
in  the  one  with  those  which  correspond  to  them  in  the  other, 
namely,  the  sides  CD  and  DB,  we  obtain  AD/ CD  =  CD/DB, 
which  was  to  be  proved. 

163.  Corollary  2.  If,  in  any  right  triangle,  a  perpendicular  is 
drawn  from  the  vertex  of  the  right  angle  to  the  hypotenuse,  each 
side  of  the  right  triangle  is  the  mean  proportional  between  the 
hypotenuse  and  the  segment  adjacent  to  that  side. 

Proof.  The  A  ADC  and  CDB  are  each  similar  to  A  ABC. 
Hence       AB/AC  =  AC/ AD  and  AB/CB  =  CB/DB. 


HI,  §  164]  SIMILAR  TRIANGLES  145 

164.  Problem  3.  To  find  the  mean  proportional  he- 
tween  two  lines. 

Given  the  two  lines  a  and  h. 

To  construct  their  mean  propor- 
tional ;  that  is,  to  con- 
struct a  line  x  such    '             ■ 
that  a/x  =  xfh.  ' • 

Construction.  Draw  a  line  of  in- 
definite extent  AB^  and  from  A  lay- 
off AC=a.    From  the  point   C  lay  off   CD  =  h  (Fig.  119). 

On  AD  as  a  diameter  draw  a  semicircle. 

At  C  erect  CR  ±  AB,  meeting  the  semicircle  at  E. 

Then  CE  is  the  required  line  x. 

Proof  Draw  AE  and  DE.  Then  /.  AED  is  a  right  angle. 
Why  ?     Complete  the  proof  by  means  of  §  162. 

EXERCISES 

1.  Find  the  mean  proportional  between  9  and  4.  Answer 
the  same  question  for  1  and  3.  What  essential  difference  exists 
between  the  kinds  of  answers  in  the  two  cases  ?     (See  §  128.) 

2.  The  altitude  drawn  to  the  hypotenuse  of  a  certain  right- 
angled  triangle  divides  the  hypotenuse  into  segments  which  are 
of  length  16  in.  and  4  in.,  respectively.  How  long  is  the 
altitude  ?  Answer  the  same  question  when  the  segments  are 
each  10  in.  long ;  also  when  they  are  respectively  15  in. 
and  5  in. 

3.  If  a  perpendicular  is  drawn  from  any  point  on  a  circle  to 
a  diameter,  what  relation  exists  between  the  perpendicular  and 
the  segments  into  which  it  divides  the  diameter  ?     (See  §  133.) 

4.  If  one  of  the  sides  of  a  right-angled  triangle  is  three  times 
the  other,  in  what  ratio  does  the  perpendicular  divide  the 
hypotenuse  ?  Ans.  1 :  9 


146 


PROPORTION     SIMILARITY 


[III,  §  165 


165.  Definitions.     Similar   Polygons.      Two  polygons  are 
said  to  be  similar  when  each  may  be  decomposed  into  the  same 
number   of   triangles 
similar  each  to  each 
and  similarly  placed. 
Thus,   in    the   figure, 
the  polygons  ABCDE 
and   A'B'C'D'E'    are 
similar.     For,   if   we 
draw  the   lines   DA, 
DB,       DAI,       D'B', 
ABCDE    is    decom- 
posed into  three  triangles  that  are  similar  to  the  three  triangles 
into  which  A'B'O'D'E'  is  decomposed. 

A  regular  polygon  has  been  defined  in  §  96  as  one  which  is 
both  equilateral  and  equiangular. 

166.  Theorem  IX.     Regular   ^polygons   of  the  same 
mimber  of  sides  are  similar. 

Given  the  two   poly- 


gons     ABCDE 
AB'C'D'E'. 


and 


To  prove  that  ABCDE 
^  A!B'C'D'E'. 

Proof.  Through  D 
draw  DA,  DB,  and 
through  U  draw  D'A!, 
D'B',   thus    dividing   each   polygon   into   three   triangles. 

Then  the  A  DEA  and  D'E'A'  are  both  isosceles,  Why? 

and  they  have  the  equal  vertex  angles  at  E  and  E',  Given 

Therefore  A  DEA  -  A  D'E'A!.      (Compare  Ex.  2,  p.  142.) 

Now  prove  that  A  DAB  ~  A  D'A'B'  by  showing  that  Z  DAB 
=  Z  D'A'B'  and  that  AD/A'D'  =  AB/A!B',  To  do  this,  first 
sliow  that  Z  EAD  =  Z  E'AHy, 


Ill,  §  167]  SIMILAR  POLYGONS  147 

167 .  Theorem  X .  The  perimeters  of  two  similar  poly- 
gons  are  to  each  other  in  the  same  ratio  as  any  two  cor- 
responding sides. 

Given  the  similar  polygons  ABCDE,A!B'C'D'E'.  See  Fig.  120 

To  prove  that 

perimeter  of  ABODE    ^  AB 
perimeter  of  ^'5'0'Zy^'      A'B'' 
Proof.     We  have 

AB^BG^^CD^DE^  Given 

A'B'     B'C     CD'     D'E'' 
Therefore 

AB  +  BC-^-CD  +  DE     ^  AB  ,     rp.    jr   .  . 44 
A'B'  +  B'G'+C'D'  +  D'E'     A'B"         '     '^ 

that  is  perimeter  of  ABODE    _  AB 

'  perimeter  of  A'B'  G'D'E'  ~  A'B' ' 

EXERCISES 

1.  Give  the  proof  of  Theorem  IX  for  a  regular  hexagon ;  for 
a  regular  octagon. 

2  In  two  similar  polygons  two  corresponding  sides  are  15 
and  20  in.,  respectively.  If  the  perimeter  of  the  first  is 
5  feet,  what  is  the  perimeter  of  the  second  ?  Ayis.  6|  ft. 

3  The  perimeters  of  two  similar  polygons  are  to  each  other 
as  5 . 8.  A  side  of  the  first  is  1  ft.  What  is  the  length  of 
the  corresponding  side  of  the  other  ? 

4.  Prove  that  the  perimeters  of  similar  polygons  are  in  the 
same  ratio  as  any  two  corresponding  diagonals  of  the  polygons. 

5.  The  perimeter  of  an  equilateral  triangle  is  3  ft.  Find 
the  side  of  an  equilateral  triangle  whose  altitude  is  one  half 
the  altitude  of  the  first  triangle.  Generalize  your  result  into 
a  theorem  relating  to  similar  triangles. 

6.  Prove  that  the  perimeters  of  similar  triangles  are  to  each 
other  in  the  same  ratio  as  any  two  corresponding  medians. 


148 


PROPORTION     SIMILARITY 


[III,  §  168 


PART   IV.     PROPORTIONAL  PROPERTIES  OF 
CHORDS,   SECANTS,  AND   TANGENTS 

168.  Theorem  XI.  If  two  chords  intersect  icitliin  a 
circle,  the  product  of  the  segments  of  the  one  is  equal  to 
the  product  of  the  segments  of  the  other. 

Given  the  chords  AC  and  BD  intersect- 
ing at  K. 

To  prove  that  KA  •  KC  =  KB  -  KD. 

Proof.  Draw  AD  and  BG.  In  the 
A  AKD  and  BKC  we  have  ZD  =  AC, 
each  being  measured  by  AB/2.  §  132 

Likewise  ZA=  Z.B.  Why  ? 

Therefore  AAKD^ABCK;  Why? 
,  KA     KD 

^^"^^  ^  =  ^' 

whence  KA  •  KO  =  KB  •  KD. 


Fig.  122 


Why 


EXERCISES 

1.  A  chord  of  a  circle  is  divided  into  two  segments  of  5  in. 
and  3  in.,  respectively,  by  another  chord  one  of  whose  segments 
is  4  in.     How  long  is  the  second  chord  ?  A^is.  7  j  in. 

2.  The  greatest  distance  to  a  chord  8  in.  long  from  a  point 
on  its  intercepted  (minor)  arc  is  2  in.  What  is  the  diameter 
of  the  circle  ? 

3.  P  is  a  fixed  point  within  a  circle  through  which  (point)  a 
chord  passes.  As  the  chord  swings  round  P  as  a  pivot,  what 
can  be  said  of  the  segments  of  all  new  chords  thus  obtained  ? 

4.  If  two  circles  intersect  and  through  any  point  in  their 
common  chord  two  other  chords  be  drawn,  one  to  each  circle, 
prove  that  the  product  of  the  segments  of  one  chord  is  equal  to 
the  product  of  the  segments  of  the  other. 


Til,  §  170]      CHORDS     SECANTS     TANGENTS 


149 


169.  Theorem  XII.  If  from  a  point  without  a  circle 
a  secant  and  a  tangent  are  drawn,  the  tangent  is  the 
mean  proportional  between  the  entire  secant  and  its  exr 
terior  segment. 


Fig.  123 

To  prove  that  AC /AD  =  AD/AB. 

Outline  of  proof.     Draw  BD  and  DC. 

Then                        Z.A=  ^A,  Hen. 

and  /.C  =  Z.  ADB,  since  each  is  measured  by  ^  DB.  Why  ? 

Therefore          A  ACD  ~  A  ABD,  §  155 

and  hence             AC/ AD  =  AD/AB.  Why  ? 

Note.  If  in  Fig.  123,  AC  is  allowed  to  swing  around  A  as 
a  pivot,  the  tangent  AD  meanwhile  remaining  fixed,  we  have 
throughout  the  motion,  AC  •  AB  =  AD  .  This,  in  fact,  is  what 
Theorem  XII  says.  But,  since  AD  remains  fixed,  this  means 
that  the  product  AG  •  AB  remians  constant  throughout  the 
motion.     This  fact  may  be  stated  as  follows: 


170.  Theorem  XIII.  If  from  a  fixed 
point  without  a  circle  any  two  secants 
are  drawn,  the  j^f^oduct  of  one  secant 
and  its  external  segment  is  equal  to  the 
product  of  the  other  secant  and  its  ex- 
ternal segment. 


Fig.  124 


150 


PROPORTION     SIMILARITY  [III,  §  170 


EXERCISES 

1.  A  secant  is  drawn  from  a  point  without  a  circle  in  such 
a  way  that  its  whole  length  is  9  in.,  while  the  part  cut  off 
within  the  circle  is  4  in.  What  is  the  length  of  the  tangent 
to  the  circle  from  the  same  point  ?  Ans.  3  V5  in. 

2.  Two  secants  are  drawn  from  the  same  point  without  a 
circle.  One  is  11  in.  long  and  its  external  segment  is  4  in. 
The  external  segment  of  the  second  is  2  in.  What  is  the 
length  of  the  second  secant  ? 

3.  One  of  two  secants  meeting  without  a  circle  is  18  in. 
long,  while  its  external  segment  is  4  in.  The  other  secant  is 
divided  into  equal  parts  by  the  circumference.  Eind  the 
length  of  the  second  secant. 

4.  A  point  P  is  12  in.  from  the  center  of  a  circle  whose 
radius  is  7  in.  A  secant  is  drawn  from  P.  Find  the  product 
of  the  entire  secant  and  its  external  segment. 

Pj 

5.  Prove  that  tangents  drawn  to  two 
intersecting  circles  from  a  point  on  their 
common  chord  are  equal.  0 

6.  Given  two  non-intersecting 
circles  0  and  0',  draw  any  circle 
ABCD  intersecting  both  of  them. 
Draw  the  two  common  chords 
AB  and  CD,  and  extend  them  to 
meet  at  P.  Show  that  tangents 
drawn  from  P  to  the  two  circles 
0  and  0'  are  equal. 

7.  Show  that  the  diameter  of  a  circular  porch  column  can 
be  found  by  measuring  certain  lines  entirely  outside  the  column. 
Can  this  still  be  done  if  over  half  the  face  of  the  column  is 
buried  in  cement  ? 


Ill,  §  172]      CHORDS     SECANTS     TANGENTS  151 

171.  Definition.  A  line  segment  is  said  to  be  divided  in 
extreme  and  mean  ratio  when  one  of  its  segments  is  a  mean  pro- 
portional between  the  whole  line  and  ^ 

the  other  segment.     In  Fig.  125,  C  is    ^'  '  '^ 

so  located  that  AB/AC  =  AC/CB,  so  ^^<^-  ^^ 

that  AB  is  divided  by  G  in  extreme  and  mean  ratio. 

Note.  The  division  of  a  line  in  extreme  and  mean  ratio  has 
been  called  the  golden  section  or  division  in  golden  mean.  It 
•was  observed  by  the  ancients  that  artistic  effects  frequently 
result  from  its  use.  Thus,  a  rectangular  picture  frame  will 
usually  give  the  best  effect  if  its  length  and  width  are  in  the 
ratio  just  described;  that  is,  in  the  ratio  of  AC  to  CB  in  Fig. 
125.  A  similar  remark  applies  to  the  height  of  the  back  of  a 
chair  as  compared  with  the  length  of  its  legs. 

172.  Problem  4.     To  divide  a  given  line  segment  in 

extreme  and  mean  ratio.  \ 

\d 

Given  the  line  A B.  i  r     -'''  > 

Required  to  divide  AB  in  extreme  \l       ^^.-^'^' 

and  mean  ratio;   that  is,  to  deter-  l-V 

mine  a  point  F  such  that  AB/AF  =       ^^^"^        \  ^ v^ 


y 


AF/FB.  A  F 

Construction.   At  BdTAwBC±AB  Fig.  126 

and  equal  to  half  of  AB. 

With  C  as  center  and  CB  as  radius  draw  a  circle. 

Draw  AC  meeting  the  circle  at  E  and  D. 

On  AB  take  AF  =:  AE. 

Then  F  is  the  point  of  division  required. 

Proof.  AD/AB  =  AB/AE ;  Why  ? 

hence  4P^^^  =  ^B-A^.  why? 

AB  AE  ^ 

But  AB  =  ED,  and  AF=AE',  Why  ? 

therefore   AF/AB  =  FB/AF,  or  AB/AF  =  AF/FB.      Why  ? 


152 


PROPORTION      SIMILARITY 


[HI,  §  173 


PART   V.     SIMILAR   RIGHT   TRIANGLES. 
TRIGONOMETRIC   RATIOS 

173.  Similar  Right  Triangles.  The  importance  of  a  special 
study  of  similar  right  triangles  results  from  the  simplicity  and 
usefulness  of  the  result,  §  157,  that  two  right  triangles  are  similar 
if  an  acute  angle  of  one  is  equal  to  an  acute  angle  of  the  other. 

This  proposition  may  now  be  restated  as  follows : 

174.  Theorem  XIV.  If  two  right  triangles  have  an  acute 
angle  of  one  equal  to  an  acute  angle  of  the  other,  their  corre- 
sponding sides  are  in  the  same  ratios. 

B 


Fig.  127 

Given  the  rt.  A  ABC  and  A'B'C,  with  ZA  =  ZA\ 
To  prove  that  their  corresponding  sides  are  in  the  same  ratios. 
Proof.  A  ABC  ~  A  A'B'C^ ;  §  157 

hence  the  corresponding  sides  are  in  the  same  ratios ; 

BC^B^,    Aa^A;C^^    BC^B^        J.   .... 

AB     A'B"    AB     A'B"    AC     A'C' 


that  is  : 


175.  Corollary  1.  If  an  acute  angle  of  a  right  triangle  is 
known,  the  ratios  of  the  sides  are  all  determined, 

176.  Corollary  2.  If  the  ratio  of  any  pair  of  sides  of  a  right 
triangle  is  given,  the  acute  angles  are  determined. 

Note.  The  study  of  the  use  of  these  ratios  is  called 
Trigonometry.  By  their  use,  any  part  of  a  right  triangle  can 
be  found,  if  any  two  parts,  not  both  angles,  are  given,  besides 
the  right  angle.     This  process  is  called  solving  the  triangle. 


Ill,  §  178]  TRIGONOMETRIC   RATIOS  153 

177.  Definitions.  Let  Z  A,  Fig.  128,  be  a  known  acute  angle 
of  a  right  triangle,  in  which  Z  C  is  a  right  angle ;  and  let  us 
denote  the  sides  opposite  ZA,ZBf 
Z  C,  by  a,  h,  c,  respectively. 

Then  the  ratios  a/b,  a/c,  b/c  are 
all  determined,  by  §  175.  These 
ratios  are  named  as  follows : 

(1)  a/c  is  called  the  sine  of  the 
angled;  Fig.  128 

(2)  b/c  is  called  the  cosine  of  the  angle  A ; 

(3)  a/b  is  called  the  tangent  of  the  angle  A. 
That  is,  for  an  acute  angle  of  a  right  triangle : 

(1)  the  sine  of  the  angle  =  side  opposite  it  -=-  hypotenuse ; 

(2)  the  cosine  of  the  angle  =  side  adjacent  it  -f-  hypotenuse ; 

(3)  the  tangent  of  the  angle  =  side  opposite  -;-  side  adjacent. 
These  ratios  are  called  the  trigonometric  ratios. 

178.  Values  of  the  Ratios.  Table.  The  values  of  the 
three  ratios  mentioned  in  §  177  can  be  obtained,  approximately, 
from  a  good  figure,  for  any  acute  angle  A. 

These  values  are  tabulated  on  p.  155,  to  three  places  of  deci- 
mals, for  every  angle  from  0°  to  90°,  at  intervals  of  1°.  The 
student  may  check  the  accuracy  of  any  entry  in  this  table  by 
drawing  an  accurate  figure  with  a  protractor,  actually  measur- 
ing the  sides  of  it,  and  then  calculating  their  ratios. 

Example.  The  length  of  the  shadow  of  a  tree  CB  is  44  ft. 
when  the  angle  CAB  between  the  shadow  CA  and  the  line  AB 
is  31°.     Find  the  height  of  the  tree. 

[Solution.  Since  BC/AC  is  the  tangent  of 
31",  by  §  177,  we  look  in  the  table,  in  the  col- 
umn headed  tangent,  and  opposite  31°.  This 
gives  the  value  .601 ;  hence  BC/AC=  Ml ;  but 
^O  =  44  ft. ;  it  follows  that  ^**'^44  ft. 

BC  =  AC  X  .601  =  44  X  .601  =  26.444  (ft.).] 


154  PROPORTION      SIMILARITY  [III,  §  178 

EXERCISES 

1.  How  large,  in  degrees,  is  an  acute  angle  whose  tangent 
isl?  ^B 

2.  The  shadow  of  a  tree  is  26  ft.  long 
when  the  angle  of  elevation  of  the  sun  (Z  CAB 
in  the  figure)  is  45°.     How  tall  is  the  tree  ? 

3.  One  side  of  a  right  triangle  is  2  in.  ; 
the  adjacent  angle  is  42° ;  determine  the  re-  ^  ^** 
maining  side  and  the  hypotenuse,  and  check  by  measurement 
from  an  accurate  figure.       Ans.  side  =  1.8  in. ;  hyp.  =  2.69  in. 

^'4.  One  side  of  a  right  triangle  is  2  in.  and  the  opposite 
angle  is  42°;  determine  the  remaining  side  and  hypotenuse. 
Check  by  measurement  in  a  figure. 

"^5.  The  hypotenuse  of  a  right  triangle  is  28  in. ;  one  angle  is 
32°.     Determine  the  two  perpendicular  sides.     Check. 

^'.  To  determine  the  height  of  a  tree  OA  standing  in  a  level 
field  the  distance  OB  =  100  ft.  from  the  base  0  of  the  tree  to  a 
point  B  in  the  field  is  measured,  and  the  angle  OBA  is  then 
found  to  be  37°.  Find  approximately  the  height  by  measure- 
ment in  a  reduced  figure,  and  by  the  table,  p.  155. 

7.  Measure  two  adjacent  edges  of  your  study  table.  Find 
the  angles  that  the  diagonal  makes  with  the  edges,  (a)  by 
drawing  an  accurate  figure  and  measuring  the  angle  with  a 
protractor ;  (b)  by  use  of  the  table  (p.  155). 

8.  The  tread  of  a  step  on  a  certain  stairway  is  10  in.  wide ; 
the  step  rises  7  in.  above  the  next  lower  step.  Find  the  angle 
at  which  the  stairway  rises,  (a)  by  a  protractor  from  an 
accurate  figure ;  (6)  from  the  table,  p.  155.     Ans.  35°  (nearly). 

9.  What  angle  does  a  rafter  make  with  the  plate  beam 
(Fig.,  Ex.  2,  p.  43),  if  the  roof  is  "half-pitch  " ;  if  the  pitch  of 
the  roof  is  1/3.  Ans.  45°;  nearly  34°. 

[The  pUch  of  a  roof  is  the  rise  divided  by  the  entire  span.] 


Ill,  §  178] 


TRIGONOMETRIC   RATIOS 


155 


[The  abbreviation  hyp  means  hypotenuse  ;  ad)  means  the  aide  adjacent  to  the  angle ; 
opp  means  the  aide  oppoitite  the  angle.    See  also  the  larger  table,  p.  ili.] 


Anglb 

Sine 

C08INK 

Tangent 

Angle 

Sink 

Cosine 

Tangent 

(opp/hyp) 

(adj/hyp) 

(opp/adj) 

(opp/hyp) 

(adj/hyp) 

(opp/adj) 

0° 

.000 

1.000 

.000 

45° 

.707 

.707 

1.000 

1° 

.017 

1.000 

.017 

46° 

.719 

.695 

1.036 

2° 

.035 

.999 

.035 

47° 

.731 

.682 

1.072 

3° 

.a')2 

.999 

.052 

48° 

.743 

.()()9 

1.111 

40 

.070 

.998 

.070 

49° 

.755 

.65(3 

1.150 

6° 

.087 

.9^)6 

.087 

50° 

.766 

.643 

1.192 

()° 

.105 

.9^)5 

.105 

51° 

.777 

.629 

1.235 

7° 

.122 

.993 

.123 

52° 

.788 

.616 

1.280 

8° 

.139 

.990 

.141 

53° 

.799 

Am 

1.327 

i>° 

.\m 

.988 

.158 

54° 

.809 

.588 

1.37(5 

10^ 

.174 

.985 

.176 

56° 

.819 

.574 

1.428 

11° 

.191 

.982 

.194 

56° 

.829 

.559 

1483 

12"^ 

.208 

.978 

.213 

57° 

.839 

.545 

1.540 

13° 

.225 

.974 

.231 

58° 

.848 

.530 

1.(500 

14° 

.242 

.970 

.249 

59° 

.857 

.515 

1.(5(54 

15° 

.259 

.966 

.268 

60° 

.86(5 

.500 

1.732 

16° 

.276 

.961 

.287 

61° 

.875 

.485 

1.804 

17° 

.292 

956 

.306 

62° 

.883 

.469 

1.881 

18° 

.309 

.951 

.325 

()3° 

.891 

.454 

l.i)63 

19° 

.326 

.946 

.^4 

64° 

.899 

.438 

2.050 

20° 

.342 

.940 

.3<>4 

65° 

.906 

.423 

2.145 

21° 

.358 

.934 

.384 

()6° 

.914 

.407 

2.246 

22° 

.375 

.927 

.404 

67° 

.921 

.391 

2.356 

23° 

.391 

.921 

.424 

68° 

.927 

.375 

2.475 

24° 

.407 

.914 

.445 

69° 

.9M 

.358 

2.605 

26° 

.423 

.906 

.466 

70° 

.910 

.342 

2.747 

26° 

.4:38 

.899 

.488 

71° 

.94f) 

.326 

2.iX)4 

27° 

.454 

.891 

.510 

72° 

.951 

.309 

3.078 

28° 

.469 

.883 

.532 

73° 

.956 

.2«)2 

3.271 

29° 

.485 

.875 

.554 

74° 

.961 

.276 

3.487 

80° 

.500 

.86() 

.577 

75° 

.9(;<) 

.259 

3.732 

31° 

.515 

.857 

.601 

76° 

.970 

.242 

4.011 

32° 

.530 

.848 

.()25 

77° 

.974 

.225 

4.331 

33° 

.545 

.839 

.649 

78° 

.978 

.208 

4.705 

34° 

.559 

.8'J9 

.675 

79° 

.982 

.191 

5.145 

35° 

.574 

.819 

.700 

80° 

.985 

.174 

5.671 

36° 

.588 

.809 

.727 

81° 

.988 

.15(5 

6.314 

37° 

.602 

.7i)9 

.754 

82° 

.990 

.139 

7.115 

38° 

.616 

.788 

.781 

8:i° 

.«H>3 

.122 

8.144 

39° 

.629 

.777 

.810 

84° 

.995 

.105 

9.514 

40° 

.643 

.766 

.839 

85° 

.996 

.087 

11.430 

41° 

.65() 

.755 

.s(;9 

8()° 

.'.>98 

.070 

i4..'m 

42° 

.6<)9 

.743 

.900 

87° 

.\m 

.052 

19.081 

43° 

.682 

.731 

.933 

88° 

.999 

.oa^ 

28(5:36 

44° 

.695 

.719 

.966 

89^ 

1.000 

.017 

57.290 

45° 

.707 

.707 

1.000 

90° 

1.000 

.000 

156 


PROPORTION     SIMILARITY 


[in,  §  179 


179.  Theorem  XV.  Corresponding  altitudes  divide  any 
tivo  similar  triangles  into  two  corresponding  pairs  of 
similar  right  triangles. 


Fig.  129 

Given  the  two  similar  triangles  ^5(7  and  A'B'C',  and  given 
the  corresponding  altitudes  BD  and  B'D'- 

To  prove  that  A  ABD  ^  A  A'B'D',  and  A  DCB  ~  A  D'CB'. 

Outline  of  proof.  Show  that  A  ABD  ~  A  A'B'D'  by  showing 
that  Z.A  =  ZA'.     (§  §  158,  174.) 

180.  Corollary  1.  Any  two  similar  polygons  may  be  sub- 
divided into  correspondiyig  pairs  of  similar  right  triangles. 

Note.  Division  of  a  triangle  into  right  triangles  is  often 
useful. 

Example.  In  Fig.  129,  suppose  c  =  10  in.,  /.A  =  30°, 
Z.B  =  85°.     Find  Z  C,  and  sides  a  and  b. 

Solution  :  First  find  Z  (7=65°.  Then  /i=cx  cosine  of  30°=10x^  =  5. 
Hence  a  =  h  ^  sine  of  O  =  5  -^  .906  =  5.52.  Again  AD  =  c  x  cosine  of 
30^=8.66  ;  and  DC=a  X  cosine  of  65°=2.33  ;  hence  b = AD -^ DC  =10.99. 

EXERCISES 

1.  The  base  of  a  certain  isosceles  triangle  is  10  in.,  and  the 
angle  at  the  vertex  is  40°.  Find  the  size  of  one  of  the  equal 
angles ;  find  the  length  of  one  of  the  equal  sides  (a)  by  measure- 
ment, (b)  using  the  table,  p.  155. 

2.  The  diagonal  of  a  certain  rectangle  is  4  ft.  One  side  is 
2  ft.  Find  the  angle  the  diagonal  makes  with  that  side,  (a)  by 
measurement,  (b)  using  the  table,  p.  155. 


Ill,  §180]         MISCELLANEOUS  EXERCISES  157 

MISCELLANEOUS   EXERCISES.     CHAPTER   III 

[The  Exercises  that  involve  the  iise  of  the  trigonometric  ratios  are 
starred,  *  ] 

^1.  A  building  casts  a  shadow  64  ft.  long.  A  lamp-post  8  ft. 
high  at  one  corner  of  the  building  casts  at  the  same  time  a 
shadow  9  ft.  long.     How  high  is  the  building  ? 

2.  Show  how  to  find  three  fifths  of  a  given  line ;  five 
sevenths. 

"  3.  A  circle  is  inscribed  in  an  isosceles  triangle.  Prove  that 
the  triangle  formed  by  joining  the  points  of  contact  is  also 
isosceles. 

4.  If  each  of  two  polygons  is  similar  to  a  third  polygon, 
they  are  similar  to  each  other.    Prove  this  statement. 

"^5.  Prove  that  in  an  inscribed  quadrilateral  the  product  of 
the  segments  of  one  diagonal  is  equal  to  the  product  of  the  seg- 
ments of  the  other. 

6.  In  the  adjoining  figure,  AB  rep- 
resents a  vertical  pole,  and  CD  a  ver- 
tical stake,  so  that  D  and  B  are  in  the 
same  line  of  sight  from  a  point  E  on 
the  level  ground  ECA.  What  measurements  must  be  taken 
to  find  ^^? 

7.  If  the  non-parallel  sides  of  a  trapezoid  are  extended  to 
meet  in  a  point  P,  the  lengths  of  the  extensions  are  propor- 
tional to  the  lengths  of  the  original  sides.     State  a  proof. 

8.  If  in  any  triangle,  a  line  is  drawn  parallel  to  the  base, 
any  line  through  the  vertex  divides  the  base  and  the  parallel 
into  segments  that  are  in  the  same  ratio.     State  a  proof. 

9.  Show  that  any  two  altitudes  h  and  h'  of  a  triangle  are 
inversely  proportional  to  the  sides  a  and  a'  to  which  they  are 
perpendicular ;  that  is,  h/h'  =  a' /a. 


158 


PROPORTION      SIMILARITY  [III,  §  180 


10.  Let  AE  and  BD  be  the  tangents  at  the  ends  of  a  diam- 
eter AB  of  a  circle  0.  Draw  any  line 
through  A,  and  suppose  it  cuts  the  circle 
at  C  and  meets  BD  at  D.  Let  E  be  the 
point  of  intersection  of  AE  and  BC  pro- 
duced. Show  that  three  similar  right  tri- 
angles are  formed;  and  show  that  AB  is 
a  mean  proportional  between  AE  and  BD. 

11.  Construct  a  fourth  proportional  to 
three  lines,  3  in.,  5  in.,  and  1  in.  long, 
respectively.     Show  that  the  resulting  line  must  be  |  in.  long. 

12.  In  general,  construct  a  fourth  proportional  to  two  given 
line  segments  a  and  b  and  the  unit  line  segment.  Show  that 
the  resulting  line  has  a  length  b/a  in  terms  of  the  unit.  [Geo- 
metric construction  for  division.'] 

13.  Show  how  to  construct  a  circle  through  two  given  points 
and  tangent  to  a  given  line.  D         c 

[Hint.  Draw  a  line  through  the  given  points  A 
and  5,  meeting  the  given  line  at  C.  Then  find  a 
mean  proportional  CD  between  AC  and  BC.  Two 
circles  are  possible  according  as  CD  is  laid  off  from  C 
in  one  direction  or  the  other  along  the  given  line.] 

14.  Show  how  to  constuct  a  circle 
tangent  to  two  given  lines  and  pass- 
ing through  a  given  point  P. 


[Hint.     Draw  BD  bisecting  Z  ABC. 
Draw  through  the  given  point  P  a  line 
PD  ±  BD  and  extend  it  to  E^   making 
DE  =  PD.     Now  apply  Ex.  13  to  draw  a  circle  through  P  and  E  and 
tangent  to  AB.     How  many  solutions  are  possible  ?] 

15.  Show  how  to  inscribe  in  a  given  circle  a  triangle  similar 
to  a  given  triangle. 

16.  Show  how  to  circumscribe  about  a  given  circle  a  triangle 
similar  to  a  given  triangle. 


Ill,  §  180]         MISCELLANEOUS  EXERCISES  159 

17.  Construct  a  fourth  proportional  to  three  lines,  1  in., 
2.5  in.,  and  3.5  in.  long.  Show  that  the  resulting  line  must 
be  2.5  X  3.5  in.  long. 

18.  Give  a  geometric  construction  for  multiplying  any  two 
numbers, 

[Hint.     It  x  —  ah,  then  1/a  =  6/ic.] 

19.  Give  a  geometnc  construction  for  enlarging  any  line 
segment  in  the  ratio  5  :  7.  Use  this  construction  to  draw  a  tri- 
angle similar  to  a  given  triangle,  with  its  sides  enlarged  5  :  7. 

20.  Show  how  to  enlarge  (or  reduce)  any  line  segment  in 
the  ratio  of  two  given  numbers  (or  two  given  line  segments). 

21.  Show  how  to  construct,  on  a  given  line  segment  as  one 
side,  a  polygon  similar  to  a  given  polygon. 

22.  Show  that  if  two  circles  are  tangent  externally,  any  line 
through  their  point  of  tangency  forms  chords  of  the  two  circles 
that  are  proportional  to  their  radii.  State  and  prove  the  anal- 
ogous theorem  when  the  circles  are  tangent  internally. 

23.  In  order  to  find  the  distance  between  two  islands  A 
and  5  in  a  lake,  what  distances  and 
angles  must  be  measured  in  the  ad- 
joining figure  ?  Compare  Ex.  6,  p. 
142.  Show  that  the  figure  can  be 
extended,  so  as  to  find  AB  by  measur- 
ing distances  only. 

^4.  If  a  line  is  drawn  parallel  to  the  parallel  bases  of  a 
trapezoid,  show  that  the  segment  cut  off  on  it  between  one  side 
and  one  diagonal  is  equal  to  that  cut  off  by  the  other  side  and 
the  other  diagonal. 

25.  The  bases  of  a  trapezoid  are  10  in.  and  15  in.  long, 
respectively,  and  the  altitude  is  8  in.  Find  the  altitude  of  the 
triangle  formed  on  the  smaller  base  by  extending  the  non- 
parallel  sides  until  they  meet.  Ans.   16  in.    . 


160  PROPORTION      SIMILARITY  [III,  §  180 

26.  Construct  a  mean  proportional  to  two  lines  3  in.  and 

4  in.  long,  respectively.  Construct  a  mean  proportional  to  two 
lines,  each  of  which  is  2  in.  long. 

27.  Construct  a  mean  proportional  to  two  lines  1  in.  and 

5  in.  long,  respectively.     How  long  is  it  ? 

v^8.  Show  how  to  construct  the  square  root  of  any  given 
number  n  by  finding  the  mean  proportional  between  a  line  of 
unit  length  and  a  line  n  units  long.  \_OeometriG  construction 
for  square  rootJ] 

29.  The  total  length  of  a  certain  secant  drawn  from  a  point 
P  to  a  circle  is  10  in. ;  its  external  segment  is  4  in.  Find  the 
length  of  the  tangent  drawn  from  P. 

30.  Taking  the  radius  B  of  the  earth  as  4000  mi.,  find  how 
far  from  a  lighthouse  150  ft.  high  the  light  is  visible.  A  closer 
value  of  R  is  3963  mi. ;  is  the  answer  changed  seriously  by 
using  this  more  accurate  value  of  E? 

w  =*  31.  If  one  side  of  a  right  triangle  is  double  the  other,  in 
what  ratio  is  the  hypotenuse  divided  by  the  altitude  drawn  to 
it  ?  What  are  the  angles  of  the  triangle  ? 
\/32.  If  one  chord  of  a  circle  is  bisected  by  another,  show 
that  either  segment  of  the  first  is  a  mean  proportional  between 
the  segments  of  the  other. 

33.  The  greatest  distance  from  a  chord  10  in.  long  to  its 
intercepted  arc  is  3  in.     Find  the  radius  of  the  circle. 

34.  A  curved  pane  of  glass  to  fit  a  window  in  a  round  tower 
is  bent  in  the  arc  of  a  circle.  If  the  width  of  the  frame  is 
30  in.  and  if  the  greatest  distance  from  the  glass  to  a  horizontal 
line  joining  its  edges  is  1.5  in.,  find  the  radius  of  the  arc. 

*  35.  The  radius  of  a  circle  is  7  ft.  What  angle  will  a  chord 
of  the  circle  11  ft.  long  subtend  at  the  center?  Check  by 
measurement  in  a  reduced  figure. 

36.  A  railroad  curve  is  to  have  a  radius  of  250  ft.  What 
is  the  greatest  distance  from  the  track  to  a  chord  100  ft.  long  ? 


Ill,  §  180]         MISCELLANEOUS  EXERCISES  161 

37.  A  railroad  curve  turns  in  the  arc  of  a  circle ;  the  greatest 
distance  from  the  track  to  a  chord  100  ft.  long  is  7  ft. ;  find  the 
radius  of  the  arc.  Aiis.  182+  ft- 

*  38.  The  width  of  the  gable  of  a  house  is  34  ft.  The  height 
of  the  house  above  the  eaves  is  15  ft. ;  find  the  length  of  the 
rafters  and  the  angle  of  inclination  of  the  roof.  Find  the  pitch 
of  the  roof.     See  Ex.  2,  p.  43,  and  Ex.  9,  p.  164. 

*  39.  To  find  the  distance  across  a  lake 
between  two  points  A  and  B,  a  surveyor 
measured  off  80  ft.  on  a  line  AC  perpen- 
dicular to  AB ;  he  then  found  Z  ACB=46°. 
Find  AB. 

*  40.  A  kite  string  is  250  ft.  long  and 
makes   an   angle   of  40°   with   the   level 

ground.     Find  (approximately)  the  height  of  the  kite  above 
the  ground,  neglecting  the  sag  in  the  string. 

*  41.  The  shadow  of  a  vertical  10-foot  pole  is  14  ft.  long. 
What  is  the  angle  of  elevation  of  the  sun  ?    Ans.  About  35.5°. 

*  42.  A  chord  of  a  circle  is  21.5  ft.  long,  and  the  angle  which 
it  subtends  at  the  center  is  41°.  Find  the  radius  of  the  circle. 
V  *  43.  The  base  of  an  isosceles  triangle  is  324  ft.,  the  angle  at 
the  vertex  is  64°  40'.     Find  the  equal  sides  and  the  altitude. 

*  44.  The  base  of  an  isosceles  triangle  is  245.5  and  each  of 
the  base  angles  is  68°  22'.  Find  the  equal  sides  and  the 
altitude. 

V*  45.   The  altitude  of  an  isosceles  triangle  is  32.2  and  each  of 
the  base  angles  is  32°  42' ;  find  the  sides  of  the  triangle. 

f  46.  Find  the  length  of  a  side  of  an  equilateral  triangle  cir- 
cumscribed about  a  circle  of  radius  15  in. 

*  47.  Show  that  the  sine  of  the  angle  A  in  Fig.  129  is  h  -^  c, 
and  that  the  sine  of  C  is  ^  -f-  a.  Hence  show  that  the  sines  of 
the  angles  A  and  C  are  proportional  to  the  sides  a  and  c  opposite 
them.  [Sine  Law.] 


CHAPTER   lY 

AREAS  OF  POLYGONS.     PYTHAGOREAN  THEOREM 

181.  Area  of  a  Rectangle.  The  fundamental  principle,  men- 
tioned in  the  Introduction  (§  25),  that  the  area  of  a  rectangle  is 
equal  to  the  product  of  its  base  by  its  height,  will  be  presupposed 
in  what  follows  in  the  present  chapter. 

The  principle  states  that  if  the  base  and  the  height  of  a  rectangle 
are,  respectively,  a  and  b,  then  its  area  (in  terms  of  the  correspond- 
ing square  unit)  is  a  •  b. 
'  The  following  corollaries  result  from  this  principle : 

182.  Corollary  1.  The  area  of  a  square  is  equal  to  the  square 
of  its  side. 

183.  Corollary  2.  The  areas  of  two  rectangles  are  to  each 
other  as  the  products  of  their  bases  and  altitudes. 

184.  Corollary  3.  Two  rectangles  that  have  equal  altitudes  are 
to  each  other  as  their  bases;  two  rectangles  that  have  equal  bases 
are  to  each  other  as  their  altitudes. 

185.  Definition.  Whenever  two  geometric  figures  have  the 
same  area  they  are  said  to  be  equal  in  area,  or  equivalent.  The 
equality  in  area  of  two  figures  is  denoted  by  the  symbol  =. 
Thus,  the  equation  AABC  =  AA'B'C  means  that  the  two 
triangles  have  equal  areas. 

EXERCISES 

1.  How  many  tiles  each  8  in.  square  will  it  take  to  tile  a 
floor  30  ft.  long  and  18  ft.  wide  ? 

2.  On  a  sheet  of  squared  paper  ruled  in  tenths  of  an  inch 
(see  p.  23),  how  many  small  squares  are  there  in  one  square 
inch? 

162 


IV,  §  185] 


RECTANGLES 


163 


3.  Compare  the  areas  of  two  rectangles  whose  altitudes  are 
equal  but  whose  bases  are  respectively  10  in.  and  7  in. 

4.  The  area  of  a  rectangle  is  400  sq.  ft. 
and  its  altitude  is  20  ft.  What  is  the  alti- 
tude of  another  rectangle  having  the  same 
base  but  whose  area  is  300  sq.  ft  ? 

5.  The  accompanying  figure  represents 
the  cross  section  of  a  steel  beam.    The  dimen- 


192,  ^  =  8.     Find 


sions  in  millimeters  are :  6=96,  ^(;=12,  h 
the  area  of  the  cross  section. 

6.  Show  that  the  adjoining  figure  il- 
lustrates the  algebraic  identity 

a  (b  +  c)  =  ah  +  ac. 
t 

7.  Show  that  the   following  figures 

illustrate  the  algebraic  identities 

(1)  (a-\-b)(a-h)  =  a'-b'',  (2)  (a -i- by  =  a^ -{- 2  ab -{- b^  • 
(3)  (a-by  =  a^-2ab-^b\ 


F 

E 

D 

o 

1 

A 

<               K 

C 

<               0 

'B^   ^-' 

L_.h J^a.b» 


I 


h^ 


-ab 


(a-b)' 


m 


(1)  (2)  (3) 

8.  Draw  a  figure  to  illustrate  each  of  the  identities 

(1)  a{})-&)=^ab-ac\    (2)  {x-\-a){x ^^^x" ■{-{a-\-b)x-^ab. 

9.  Draw  a  rectangle  5  in.  long  by  6  in.  wide.  Show  that 
either  diagonal  divides  the  rectangle  into  two  congruent  right 
triangles.     Find  the  area  of  each  of  these  triangles. 

Ana.   15  sq.  in. 

10.   Find  the  area  of  a  right  triangle  whose  two  sides  are  4 
in.  and  7  in.  long,  respectively. 


164 


AREAS 


[IV,  §  186 


186.   Theorem  I.    The  area  of  a  parallelogram  is  equal  to  the 
product  of  its  base  by  its  altitude. 


Given  the  parallelogram  ABCD  with  h  its  base  and  a  its 
altitude. 

To  prove  that  the  area  of  the  parallelogram  ABCD  is  equal 
to  a  'h. 

Proof.  Draw  AF 1.  CD  prolonged.  Then  ABGF  is  a  rec- 
tangle having  the  base  h  and  the  altitude  a. 

In  the  right  A  AFD  and  BGC  we  have  AF  =  BG  and  AD 
=  BC.  Why? 

Therefore  A  AFD  ^  A  BGC.  Why  ? 

Taking  away  A  AFD  from  the  figure  ABCF.  the  parallel- 
ogram ABCD  remains. 

Taking  away  A  BGC  from  the  same  figure,  the  rectangle 
ABGF  remains. 

Therefore  O  ABCD  =  rectangle  ABGF. 

But  rectangle  ABGF=ab,  §  181 

whence,  O  ABCD  =  ah. 

1^187.  Corollary  1.  (a)  Two  parallelograms  are  to  each  other 
as  the  prodticts  of  their  bases  and  altitudes. 

(b)  Two  parallellograms  that  have  equal  bases  and  equal  alti- 
tudes are  equal  in  area. 

I  188.  Corollary  2.  Two  parallelograms  that  have  equal  alti- 
tudes are  to  each  other  as  their  bases;  two  parallelograms  that 
have  equal  bases  are  to  each  other  as  their  altitudes. 


IV,  §  188J 


PARALLELOGRAMS 


165 


EXERCISES 

1.  In  a  certain  parallelogram  the  acute  angle  included  be- 
tween the  sides  is  30°.  If  the  base  and  altitude  are  respec- 
tively 14  in.  and  10  in.,  what  is  the  area  ?  Answer  the  same 
question  in  case  the  included  angle  is  60°. 

2.  In  the  accompanying 
figure  are  a  number  of  par- 
allelograms each  hfeving  the 
same  base  and  altitude. 
Compare  their  areas.  Does 
the  area  of  the  parallelo- 
gram depend  upon  the  angles  included  by  its  sides  ? 

3.  Prove  that  the  lines  joining  the  middle  points  of  the 
opposite  sides  of  a  parallelogram  divide  it  into  four  parallel- 
ograms that  are  equal  in  area. 

4.  Construct  a  parallelogram  double  a  given  parallelogram 
and  equiangular  to  it.     How  many  solutions  are  possible  ? 

5.  Construct  a  parallelogram  double  a  given  parallelogram 
and  having  one  of  its  angles  equal  to  a  given  angle. 

6.  What  is  the  locus  of  the  intersection  of  the  diagonals 
of  a  parallelogram  whose  base  is  fixed  and  whose  area  is 
constant  ? 

7.  ABCD  is  a  jointed  parallelogram  frame,  that  is,  it  con- 
sists of  four  pieces  of  stiff  material 
hinged  at  each  of  the  points  A,  B, 
C,  and  X),  and  such  that  side  AD 
=  side  5(7  and  side  AB  =  side  DC. 
If  the  base  AB  is  held  fixed  while 
DC  is  raised  and  lowered  into  vari- 
ous positions,  will  the  areas  of  the  various  parallelograms  be 
changing  ?  If  so,  what  will  be  the  greatest  area  obtainable  and 
what  the  least,  provided  that  AB  =  6  in.  and  BC  =  4:  in.  ? 


166  AREAS  [rV,  §189 

189.   Theorem  II.     The. area  of  a  triangle  is  equal  to  one 
half  the  product  of  its  base  and  its  altitude. 


Given  the  triangle  ABC,  having  the  base  b  and  the  altitude  h. 
To  prove  that  the  area  of  the  triangle  ABC  is  equal  to  ^  hb. 
Proof.     Construct  the  O  ABCD. 

Then  LJABCD  =  hb.  §186 

But  A  ABC  =  \n  ABCD.  §  82 

Tl^refore  AABG  =  ^hb. 

"iSO.    Corollary  1 .     (a)   Tn-o  triangles  are  to  each  other  as  tJie 
•products  of  their  bases  and  altitudes. 

(b)  Tico  triangles  that  have  equal  bases  are  to  each  other  as 
their  altitudes. 

(c)  Two  triangles  that  have  equal  altitudes  are  to  each  other  as 
their  bases. 

(d)  Tvjo  triangles  that  have  equal  bases  and,  equal  altitudes 
are  equal  in  area. 

EXERCISES 

1.  What  is  the  area  of  a  right  triangle  whose  sides  are 
respectively  3  in.  and  5  in.  ? 

2.  Draw  several  triangles  having  equal  bases  and  equal 
altitudes,  as  an  illustration  of  §  190  (d).  Are  such  triangles 
necessarily  congruent  ? 

3.  Compare  two  triangles  with  equal  altitudes  if  the  base 
of  the  tirst  is  two  thirds  that  of  the  second 


IV,  §  190]  TRIANGLES  167 

^A.  Show  that  any  median  of  a  triangle  divides  it  into  two 
equal  triangles.    Is  the  same  true  of  any  altitude  ?    Give  reason. 

*^.  What  is  the  locus  of  the  vertices  of  all  triangles  having 
a  common  base  and  the  same  area  ? 

6.  An  ordinary  elastic  rubber  band  is  stretched  out  and 
placed  around  two  pins  A  and  B  which  are  stuck  into  a  board. 
The  pins  are  placed  at  the  extrem- 
ities of  a  diameter  of  a  circle,  as 
indicated  in  the  figure.  One  of  the 
halves  of  the  band  is  now  thrust 
aside  by  means  of  a  pencil  point  so 
that  the  band  becomes  stretched  out  ^ 

into  the  form  of  an  elastic  triangle  having  one  of  its  vertices 
on  the  circle  at  C.  If  the  pencil  be  now  moved  about  on  the 
circumference,  the  band  meanwhile  slipping  over  the  point,  va- 
rious triangles  are  formed,  such  as  those  indicated  by  dotted 
lines  in  the  figure.  Do  these  triangles  all  have  the  same  area  ? 
If  not,  what  is  the  greatest  area  obtainable  for  any  triangle, 
and  what  the  least,  provided  that  the  distance  between  the 
pins  is  6  in.  ? 

n.  Show  that  the  diagonals  of  a  parallelogram  divide  it 
into  four  equal  triangles. 

8.  If  a  line  is  drawn  from  the  vertex  of  a  triangle  to  any 
point  P  in  the  base,  show  that  the  areas  of  the  two  triangles 
formed  are  to  each  other  as  the  segments  of  the  base  made  by  P. 

9.  Prove  that  if  the  middle  points  of  two  sides  of  a  tri- 
angle are  joined,  a  triangle  is  formed  whose  area  is  one  fourth 
the  area  of  the  given  triangle. 

10.  Show  that  the  area  of  a  rhombus  is  equal  to  one  half  the 
product  of  its  diagonals. 

11.  Prove  that  the  area  of  an  isosceles  right  triangle  is  equal 
to  one  fourth  of  the  area  of  the  square  erected  upon  its 
hypotenuse. 


168 


AREAS 


[IV,  §  191 


191.  Theorem  III.  The  area  of  a  trapezoid  is  equal 
to  the  product  of  its  altitude  and  one  half  the  sum  of 
its  bases. 

D 


Fig.  132 

Outline  of  Proof.     Draw  the  diagonal  BD. 
Then  A  ABD  =  ah/2  and  A  BCD  =  bh/2 ; 

hence  ABCD  =  ^  +  M  =  //^  ^ 

2       2         I     2 


Why 


192.  Corollary  1.  TTie  area  of  a  trapezoid  is  equal  to  the  prod- 
uct of  its  altitude  and  the  line  joining  the  mid-points  of  the  non- 
parallel  sides. 

D .C 


Fig.  133 

[Hint.  To  prove  area  of  ABCD  =  h  •  EF.  It  may  be  easily  shown 
that  EF  =  (AB  +  DC)/2.     See  §  89.] 

Note.  The  median  of  a  trapezoid  is  a  straight  line  that  joins 
the  middle  points  of  the  non-parallel  sides.  Thus,  in  Fig.  133, 
EF  is  the  median  of  the  trapezoid  ABCD. 

The  corollary  of  §  192  is  frequently  stated  in  the  following 
form  :  The  area  of  a  trapezoid  is  equal  to  the  product  of  its  alti- 
tude and  its  median. 


IV,  §  193] 


TRAPEZOIDS 


169 


— r 

28-: 


EXERCISES 

1.  Find  the  area  of  the  trapezoid  whose  bases  are  6  in.  and 
4  in.  respectively  and  whose  altitude  is  3  in.       Ans.  15  sq.  in. 

2.  Find  the  area  of  a  trapezoid  whose  median  is  8  in.  and 
whose  altitude  is  6  in.  i30' 

3.  An  excavation  for   a  railway 
track   is  28  ft.  deep,  130  ft.  wide 
at  the  top,  and  90  ft.  wide  at  the  so' 
bottom.     What  is  the  area  of  its  cross  section  ? 

193.  Theorem  IV.  Tivo  triangles  that  have  an  acute 
angle  of  the  one  equal  to  an  acute  angle  of  the  other  are 
to  each  other  as  the  products  of  the  sides  including  the 
equal  angles. 


Given  the  A  ABC  and  A'B'C  having  the  Z  C  common. 
A  ABC        AC'BC 


To  prove  that 

Proof.  Drawls', 
the  same  altitude  /i, 


A  A'B'C      A'CB'G 
Then,  since  the  triangles  ABC,  ^B'Ohave 


and  likewise 
Multiplying,  we  obtain 


A  ABC 

AAB'C' 
AABfC 
A  A'B'C 
A  ABC 


BG 

BCf 

AC 

A'C 

^  AC'BC 

A  A'B'C     A'C'BG 


§  190,  (c) 


170  AREAS  [IV,  §  19d 

194.   Theorem  V.     Similar  triangles  are  to  each  other 
as  the  squares  of  any  two  corresponding  sides. 


Fig.  135 
Given  the  similar  A  ABC  and  A'B'C. 


To  prove 

that 

A  ABC       AB^ 

A  A'B'C      A'W 

Proof. 

AA=Z.A. 

Therefore 

A  ABC        AC-AB 
A  A'B'C     A'C'A'B' 

But 

AC       AB 
A'C     A'B' 

Therefore 

A  ABC         AB'AB        AW 
A  A'B' C     A'B'  .  A'B'     a'B'^ 

Why? 
§  193 

Why? 


195.  Corollary  1.  The  areas  of  two  similar  polygons  are  to 
each  other  as  the  squares  of  any  tivo  corresponding  sides. 

[Hint  Divide  the  polygons  up  into  the  same  number  of  similar  trian- 
gles, as  in  Fig.  120,  §  165,  then  apply  Theorem  V,  together  with  §  144, 
Theorem  H.] 

Note.  Since  any  two  corresponding  lines  in  two  similar 
figures  are  proportional  to  two  corresponding  sides,  it  follows 
that  the  areas  of  any  tivo  similar  polygons  are  to  each  other  as 
the  squares  of  any  two  correspoyiding  lines.  Compare,  in  par 
ticular,  Exs.  4  and  6,  p.  171. 


IV,  §  195]  SIMILAR  TRIANGLES  171 

EXERCISES 

1.  Compare  the  areas  of  two  similar  triangles  whose  corre- 
sponding sides  are  in  the  ratio  3  :  4.  Ans.  9  :  16. 

2.  Draw  on  heavy  cardboard  two  triangles  whose  sides  are 
in  the  ratio  1 :  2.  Cut  these  triangles  out  and  weigh  each  of 
them.     Show  that  their  weights  should  be  in  the  ratio  1 :  4. 

3.  The  areas  of  two  similar  triangles  are  100  square  feet  and 
64  square  feet  respectively.  Compare  the  lengths  of  their  cor- 
responding sides.  Answer  the  same  question  when  the  given 
areas  are  respectively  31  square  feet  and  17  square  feet.  What 
distinction  is  to  be  made  between  the  two  cases  ? 

4.  Prove  that  the  areas  of  two  similar  triangles  are  to  each 
other  as  the  squares  of  any  two  corresponding  altitudes. 

5.  One  side  of  a  polygon  measures  8  feet  and  its  area  is 
120  square  feet.  The  corresponding  side  of  a  certain  similar 
polygon  measures  20  feet.  What  is  the  area  of  the  second 
polygon  ? 

/  6.    Prove  that  the  areas  of  any  two  similar  polygons  are  to 
each  other  as  the  squares  of  any  two  corresponding  diagonals. 

7.  If  one  squai-e  is  double  another,  what  is  the  ratio  of  their 
sides?  ^   / 

8.  In  the  figure,  BE  is  per- 
pendicular to  OD  while  BF  is 
perpendicular  to  AD.  Prove 
that  AB  'BE  =  A1)'  BF. 

9.  In  what  ratio  must  the  altitude  of  a  triangle  be  divided 
by  a  line  drawn  parallel  to  the  base  in  order  that  the  area  of  the 
triangle  may  be  divided  into  two  equal  parts  ? 

A71S.  1  :  ( V2  -  1). 

[Hint.  Call  h  llie  altitude  of  the  given  triangle  and  let  xbe  the  altitude 
of  the  small  triangle  cut  off  by  the  parallel  to  the  base.  Form  an  equation 
between  h  and  x,  solve  for  x  and  then  form  x  f  {h  —  x).] 


172 


AREAS 


fIV,  §  196 


196.  Theorem  VI.  The  Pythagorean  Theorem.  The  square 
on  the  hypotenuse  of  a  right  triangle  is  equivalent  to  the  sum  of 
the  squares  on  the  two  sides. 


Given  the  rt.  A  ABC  having  AB  as  its  hypotenuse. 


To  prove  that  AB'  =  AC  +  BC\ 

First  Proof.   Draw  CPl.  AB  and  prolong  it  to  meet  FHat  G. 

Draw  CF  and  BE. 

Since  A  ACD  and  ACB  are  rt.  A,  DGB  is  a  straight  line. 

Similarly,  AC  J  is  a  straight  line. 

In  A  ABE  and  AFC,  AF  =  AB  and  AE  =  AC.  Why  ? 

Also  Z  BAE  =  Z  CAF,  since  each  consists  of  a  right  angle 
plus  the  angle  BAC.     Therefore  A  ABE  ^  A  AFC.        Why  ? 

But  the  rectangle  AFGP  =  2  •  A  ^FC,  since  both  have  the 
same  base  AF  and  the  same  altitude  FO  (or  AP). 

Likewise,  the  square  ACDE  =  2  •  A  ABE,  since  each  has  the 
same  base  AE  and  a  common  altitude  AC. 

Therefore  rectangle  AFGP  ==  square  ACDE.  Ax.  9 

Show  similarly  that  rectangle  BPGH  =  square  CBIJ. 

Therefore,  AFGP  +  BPGH  =  ACDE  +  CBIJ.  Ax.  1 


That  is. 


AB'  =A&  +  BCf. 


IV,  §  197]  PYTHAGOREAN  THEOREM  173 

Second  Proof.     By  §  163, 

Aff  =  AB  '  AP;  and  CB"  =  AB-PB; 
hence  AC^  +  Cl?  =  AB{AP  +  PB)  =  A&. 

Note  1.  This  second  proof  is  the  same  in  principle  as  the 
first,  for  AC  —  square  ACDE,  and  AB  •  AP  =  rectangle 
AFGP  and  AB-  PB  =  rectangle  PGHB.  Eachjroof  consists 
essentially  in  showing  that  Off  =  AB  •  PB  and  AG^  =  AB  -  AP. 

The  second  proof  might  have  been  given  in  connection  with 
§163. 

Note  2.  It  should  be  observed  that  Theorem  VI,  though  re- 
lating specifically  to  areasy  furnishes  at  once  a  rule  for  finding 
the  length  of  the  hypotenuse  of  any  right  triangle  when  the 
lengths  of  its  sides  are  known.  Thus,  if  a  and  b  are  the  sides, 
the  hypotenuse  h  will  be  determined  by  the  formula  h=  ^a?-\-h'^. 
Similarly,  the  theorem  furnishes  a  rule  for  finding  either  side 
of  a  right  triangle  when  the  hypotenuse  and  the  other  side  are 
known,  the  formula  then  being  a  =  VA^  —  b\  These  two  formu- 
las are  of  great  value  in  mathematics. 

Note  3.  Aside  from  its  scientific  value.  Proposition  VI  is 
of  great  interest  historically.  Though  its  origin  is  not  known 
exactly,  it  is  supposed  to  have  been  first  proved  by  the  Greek 
mathematician  Pythagoras,  who  died  about  500  B.C.  Pythag- 
oras in  his  later  life  settled  in  Italy  and  was  identified  with 
a  group  of  mathematicians  and  philosophers  known  as  the 
Pythagorean  School.  The  school  itself  was  disrupted  after 
about  200  years,  owing  to  political  disturbances ;  but  its  influ- 
ence continued  a  strong  factor  in  the  study  apd  development 
of  mathematics. 

197.  Corollary  1.  TTie  square  on  either  side  of  a  right  triangle 
is  equivalent  to  the  square  on  the  hypotenuse  diminished  by  the 
square  on  the  other  side. 


174  AREAS  [IV,  §  197 

EXERCISES 

1.  A  baseball  diamond  is  a  square  90  ft.  on  a  side.  What 
is  the  distance  from  home  plate  to  second  base?  (Extract 
square  root  correct  to  two  decimal  places.)  Ans.  127.27  ft. 

^2.  A  ladder  30  ft.  long  is  placed  against  a  wall  with  its 
foot  8  ft.  from  the  wall.  How  far  is  the  top  of  the  ladder  from 
the  ground  ? 

3.  A  tree  is  broken  20  ft.  from  the  ground.  The  top  strikes 
the  ground  20  ft.  from  the  foot,  while  the  other  end  of  the 
broken  part  remains  attached  to  the  trunk.  How  high  was 
the  tree  ? 

*^4.  Two  columns  60  ft.  and  40  ft.  high  respectively  are  30  tt. 
apart.     What  is  the  distance  between  their  summits  ? 

5.  Show  that  it  is  possible  to  have  a  right  triangle  whose 
hypotenuse  and  two  sides  have  the  respective  values  5,  4,  3. 
In  general,  can  a  right  triangle  be  found  whose  hypotenuse 
and  two  sides  have  any  three  given  values  as  h,  a,  b  ?  If  not, 
when  is  it  possible  ? 

6.  Find  the  altitude,  and  then  the  area,  of  an  equilateral 
triangle  having  a  side  equal  to  6  in. 

7.  Find  the  area  of  an  equilateral  triangle  whose  altitude  is 
6  in. 

8.  Prove  the  theorem  stated  in  Ex.  11,  p.  167,  by  means  §  196. 

9.  Show  that  the  square  of  the  diameter  of  a  circle  is  equal 
to  the  sum  of  the  squares  of  any  two  chords  drawn  from  a 
point  on  the  circle  to  the  ends  of  the  diameter.     (§  133.) 

10.  A  kite  (see  Ex.  21,  p.  123)  is  inscribed  in  a  circle  whose 
diameter  is  24  ft.  If  the  length  of  one  of  the  two  longer  sides 
of  the  kite  is  18  ft.,  how  long  is  one  of  the  shorter  sides  ? 

11.  Show  that  the  sum  of  the  squares  of  the  four  sides  of  any 
kite  (Ex.  30,  p.  87)  is  equal  to  twice  the  sum  of  the  squares  of 
the  four  segments  of  the  cross  formed  by  its  diagonals. 


IV,  §  198] 


PYTHAGOREAN  THEOREM 


175 


12.  Show  that  the  square  of 
the  length  of  the  tangent  AP 
to  a  circle  from  a  point  P  plus 
the  square  of  the  radius  of  the 
circle  is  equal  to  the  square  of 
the  distance  OP  from  P  to  the 
center  of  the  circle. 

13.  In  the  figure  of  Ex.  12,  let  OP=p,  OA  =  r, 
Show  by  means  of  P^x.  12  that 


AP^t 


Jf  —  7 


.2_ 


(p-r)(p  +  r). 


Since  p  +  r  \s  the  length  of  the  whole  secant  from  P  through  0 
to  the  opposite  side  of  the  circle,  and  since  p  —  r  is  the  external 
segment  of  this  secant,  show  that  the  preceding  equation  also 
results  from  §  169. 

14.  In  any  right  triangle  ABC  (see  Fig.  126,  p.  151),  draw 
a  circle  with  radius  CB  about  C  as  center.  Using  the  lettering 
of  Fig.  126,  reprove  Th.  VI  by  means  of  §  169,  by  showing  that 

AB'  =  AE.AD=(AO-  EC)(AO  +  CD) 
=  (AC-CB)(AC+CB) 
^AC^-CB", 
or  AC''=AB^-]-GB\ 


198.   Definition.     The  projection  of 

a  line  AB  upon  another 

^— iB 

.8 

!                     !       A/           !            M    , 

/     i        M                   N 

M                  N       M             N            1/ 

N        j                     i       ° 

A 

1         ^^^ 

Fio.  137 


line  CD  is  the  portion  of  CD  cut  off  between  the  perpendicu- 
lars drawn  from  the  extremities  of  AB  to  CD.  Thus,  in  the 
figure,  MN  is  in  each  instance  the  projection  of  AB  upon  CD. 


176 


AREAS 


[IV,  §  199 


199.  Theorem  VII.  In  any  triangle  the  square  on  the 
side  opposite  the  acute  angle  is  equal  to  the  sum  of  the 
squares  on  the  other  tioo  sides  diminished  hy  twice 
the  product  of  one  of  those  sides  and  the  projection  of 
the  other  upon  it 
A 


K — P— ^ 
Fig.  138 

Given  the  A  ABO  in  which  C  is  an  acute  angle.  Let  a,  b,  c 
be  the  sides  opposite  the  angles  A,  B,  C  respectively  and  let 
p  represent  the  projection  of  b  upon  a. 

To  prove  that      c^  =  a^ -{- b^  —  2  ap. 

Proof.  Draw  the  altitude  h  upon  the  base  a.  Then,  in 
Fig.  138, 


c2  =  7i2  +  (a-p)2. 


52_^2_|_^2_2a^_^p2 


Why? 
Why? 
Why? 

The  re- 


But 
hence 
or  c^  =  a2  +  62  _  2  ap. 

Likewise,  in  Fig.  139,  we  have  e'^  =  h^  -{-{p  —  ay 
maining  details  in  this  case  are  left  to  the  student. 

EXERCISES 

1.  Show  that  if  b  =  c  in  Fig.  138,  a^  =  2ap,  or  a  =  2 p. 
Compare  §  43. 

2.  Show  that  if  c  =  a  in  Fig.  138,  b^  =  2  ap. 

3.  Show,  from  Ex.  2,  that  the  base  of  any  isosceles  triangle 
is  a  mean  proportional  between  one  of  the  equal  sides  and  twice 
the  projection  of  the  base  upon  it. 


IV,/§200]  PYTHAGOREAN  THEOREM  177 


200.  Theorem  VIII.  In  any  obtuse  triangle  the  square 
on  the  side  opposite  the  obtuse  angle  is  equal  to  the  sum 
of  the  squares  on  the  other  two  sides  increased  hy  twice 
the  product  of  one  of  those  sides  and  the  projection  of 
the  other  upon  it. 

Given  the  obtuse  A  ABC 
in  which  C  is  the  obtuse 
angle.  Let  a,  b,  c  be  the 
sides  opposite  the  angles 
A,  B,  C  respectively  and 
let  p  represent  the  proiec- 
tion  of  0  upon  a. 

To  prove  that        c^  =  a^  +  6^  4.  o  ap. 

[The  proof,  being  similar  to  that  of  §  199,  is  left  to  the  student.] 

EXERCISES 
,    [The  symbol  *  means  that  the  exercise  requires  the  use  of  trigono- 
metric ratios.] 

1.  How  does  the  square  on  any  side  of  a  triangle  opposite 
an  acute  angle  compare  with  the  sum  of  the  squares  on  the 
other  two  sides  ?  Answer  the  same  question  for  the  side 
opposite  the  obtuse  angle  in  an  obtuse  angled  triangle. 

2.  Show  that  either  Theorem  VII  or  VIII  when  applied 
to  a  right  triangle  gives  Theorem  VI. 

3.  Find  the  area  of  an  isosceles  triangle  whose  side  is  11  in. 
and  whose  base  is  8  in. 

\A.   Prove  that  the  sum  of  the  squares  on  the  diagonals  of  a 
parallelogram  is  equal  to  the  sum  of  the  squares  on  the  four  sides. 

*  5.  Show  that  the  projection  of  any  line  AB  on  another 
line  CD  (Fig.  137)  is  equal  to  the  product  of  that  line  and  the 
cosine  of  the  angle  between  AB  and  a  parallel  to  CD  through  A. 

*  6.  By  means  of  Ex.  5,  show  that  Theorem  VII  gives 
(^  =  a}-\-W  —  2ahx  (cosine  of  C).     [The  Cosine  Law.] 


178 


AREAS 


[IV,  §  201 


201.    Problem  1.     To  construct  a  square  ivhose  area  shall  be 
equal  to  the  sum  of  the  areas  of  tivo  given  squares. 


Given  the  squares  li  and  S  having  respectively  the  sides  r 
and  s. 

Required  to  construct  a  square  T  whose  area  shall  be  the  sum 
of  the  areas  of  B  and  S. 

Construction.     Draw  a  right  A  having  r  and  s  as  sides. 

Upon  the  hypotenuse  h  of  this  triangle  construct  a  square  T. 
This  is  the  square  desired. 

The  proof  follows  immediately  from  Theorem  VI  and  is 
therefore  left  to  the  student. 

EXERCISES 

1.  Show  how  to  construct  a  square  equal  in  area  to  the  sum 
of  three  given  squares.  Generalize  your  answer  to  the  case  of 
any  number  of  given  squares. 

2.  To  construct  a  square  equal  in  area  to  the  difference  of 
two  given  squares. 

3.  Show  that  a  square  erected  on  the  diagonal  of  a  given 
square  is  equal  to  twice  the  given  square. 

4.  Show  that  the  square  erected  on  half  the  diagonal  of  a 
square  is  equal  to  half  the  given  square. 

5.  Construct  a  square  equal  to  a  given  triangle. 

[Hint.  The  side  of  the  square  is  the  mean  proportional  between  the 
base  and  half  the  altitude  of  the  triangle.     Why  ?] 


IV,  §  202]  CONSTRUCTIONS  179 

202.   Problem  2.     To  construct  a  triangle  whose  area  shall  be 
equal  to  that  of  a  given  polygon. 

Given  the  polygon  P^ 

ABODE.  y^^T^^^ 

Required  to  construct  // '     \  ^^N^ 

a  triangle  equivalent  to         ^/ /    l         \        \J\^ 
ABODE.  f\/    j  \        \^5 

Construction.     Draw         //\    /  ^^       .y^\^ 

DA  and  through  E  draw    __y_ V 1*/^ /N» 

EF II  DA    and    meeting        FA  B  G 

BA   prolonged   at   F.  Fig.  142 

Draw  DF. 

Then  the  polygon  FBOD  has  one  side  less  than  the  polygon 
ABODE,  but  is  equivalent  to  it.     (See  proof  below.) 

This  process  may  be  continued  until  the  last  polygon  reached 
is  the  triangle  desired. 

Proof.     A  AED  =  A  AFD,  since  each  has  the  same  base  AD 
and  their  altitudes  are  equal.  §«i.90,  d 

Adding  polygon  ABOD  to  both  members  of  this  equation,  we 
obtain  Polygon  FBOD  =  Polygon  ABODE. 

Similarly  we  have  Polygon  FBOD  =  A  FGD. 

Therefore  A  FOD  is  the  triangle  required. 

EXERCISES 

1.  Construct  a  triangle  equivalent  to  a  square  whose  side  is 
2  in.     Is  your  triangle  the  only  such  triangle  ? 

2.  Construct  a  square  equivalent  to  a  given  parallelogram. 

[Hint.    The  side  of  the  square  is  the  mean  proportional  between  the 
base  and  altitude  of  the  parallelogram.     Why  ?] 

3.  How  could  a  square  be  constructed  that  would  be  equiva- 
lent to  a  given  pentagon  ? 

[Hint.     Construct  a  triangle  equal  to  the  given  pentagon ;  then  pro- 
ceed as  in  Ex.  5,  p.  178.] 


180  AREAS  [IV,  §  202 

MISCELLANEOUS   EXERCISES.      CHAPTER   IV 

[The  symbol  *  means  that  the  exercise  involves  Trigonometric  ratios.] 

1.  Find  the  area  of  a  square  whose  diagonal  is  30  ft.  long. 

2.  Find  the  difference  in  boundary  between  a  rectangle 
whose  base  is  16  ft.  and  a  square  equal  to  it  whose  side  is  12  ft. 

3.  Prove  that  the  diagonal  of  a  trapezoid  divides  it  into  two 
triangles  whose  areas  are  proportional  to  the  bases. 

4.  Show,  by  Ex.  3,  that  the  perpendiculars  let  fall  upon 
one  diagonal  of  a  trapezoid  from  the  ends  of  the  other  diagonal 
are  proportional  to  the  parallel  bases. 

5.  Show  how  to  find,  from  proper  measurements,  the  area  of 
a  vacant  lot  bounded  by  four  streets,  only  one  pair  of  which  are 
parallel. 

6.  A  field  of  the  form  of  a  right  triangle  containing  9  acres 
is  represented  on  a  map  by  a  right  triangle  whose  sides  are  17 
in.  and  25  in.  On  what  scale  is  the  plan  drawn  ?  [1  acre  =  4840 
sq.  yd!] 

7.  What  is  the  locus  of  all  points  such  that  the  sum  of  the 
squares  of  the  distances  of  any  one  of  them  from  two  fixed 
points  is  equal  to  the  square  of  the  distance  between  those  two 
points  ? 

8.  Obtain  the  formula  for  the  area  of  an  isosceles  right 
triangle  whose  hypotenuse  is  h. 

9.  Obtain  the  formula  for  the  area  of  an  isosceles  triangle 
whose  base  is  b  and  whose  side  is  a. 

10.  From  Ex.  9  derive  a  formula  for  the  area  of  an  equi- 
lateral triangle. 

11.  If  one  side  of  a  right  triangle  is  three  times  the  other, 
how  long  is  the  hypotenuse  as  compared  with  the  shorter  side  ? 

12.  In  Ex.  11,  what  are  the  lengths  of  the  segments  of  the 
hypotenuse  made  by  a  perpendicular  from  the  vertex  of  the 
right  angle,  in  terms  of  the  smaller  side  of  the  triangle  ? 


IV,  §  202] 


MISCELLANEOUS  EXERCISES 


181 


13.  A  channel  for  water  has  a  cross  section  which  is  of  the 
form  of  a  trapezoid  ABCD. 

If  AD  and  BC  are  known  and  the  total  height  H  of  the 
channel  is  known,  find  the  area  of  the 
cross  section. 

If  the  depth  of  the  water  is  h,  find 
the  area  of  the  cross  section  of  the  water 
in  the  channel ;  (1)  when  h  =  H/2 ; 
(2)  when  h  =  H/i ;  when  7i  =  2  H/S. 

Find  the  length  of  CD  in  terms  of  H  if  the  angle  ADC  is  45°. 

If  the  water  level  is  EF,  find  CF  in  terms  of  h,  if  the  angle 
ADC  is  45°. 

14.  Do  the  areas  mentioned  in  Ex.  13  change  if  BC,  H,  and 
AD  remain  fixed,  but  the  angles  at  B  and  C  are  changed  as 
represented  in  the  figure  ?     Explain  your  answer. 

15.  The  area  of  a  field  may  be  found  in  the  following  way : 
Run  (stake  out)  a  line 
(base  line)  AB.  From 
certain  points  in  the 
boundary  of  the  field  the 
distances  to  this  line  are 
measured,  as  a,  b,  c,  d,  e, 
/,  g,  and  the  distances,  j, 
k,  I,  m,  n,  0,  are  also  meas- 
ured. Complete  the  de- 
scription of  how  to  proceed. 

16.  To  lay  off  a  right  angle,  carpenters  frequently  use  three 
sticks  3  ft.,  4  ft.,  and  5  ft.  long,  respectively.  Show  that 
these  sides  form  a  right  triangle. 

17.  Show  that  any  multiples  of  3, 4,  6,  may  be  used  in  place 
of  3,  4,  5  of  Ex.  16. 

18.  Can  you  discover  any  three  integral  numbers  a,  b,  c,  not 
multiples  of  3,  4,  5,  for  which  a^  +  b^^c^? 


182 


AREAS 


[IV,  §  202 


\l9.  Prove  that  if  the  middle  point  of  one  of  the  non-parallel 
sides  of  a  trapezoid  is  joined  to  the  extremities  of  the  other 
non-parallel  side,  the  area  of  the  triangle  thus  formed  is  equal 
to  half  that  of  the  trapezoid. 

20.  TF  is  a  wall  having  a  round  corner 
upon  which  a  gutter  is  to  be  placed,  and  it 
is  desired  to  find  the  radius  of  the  circle  of 
which  ABC  is  a  quadrant.  If  the  line  AC 
measures  24  ft.,  show  that  the  desired  radius 
will  be  about  17  ft. 

21.  Find  the  diagonals  of  a  rhombus  whose  side  is  6  ft.  1  in. 
and  whose  area  is  9  sq.  ft. 

22.  Find  a  formula  for  the  altitude  h  of  an  equilateral  tri- 
angle in  terms  of  one  side  a,  and  find  the  ratio  of  this  altitude 
to  (1)  the  whole  side,  (2)  one  half  of 
one  side. 

*  23.  From  Ex.  22  show  that  the  sine 
of  60°  is  equal  to  V3/2.  Show  also 
that  the  cosine  of  60°  is  equal  to  i; 
and  that  the  tangent  of  60°  is  equal 
to  V3.  (See  §  177.)  Check  by  means 
of  the  table,  p.  155. 

*  24.  By  means  of  an  isosceles  right  triangle,  show  that  the 
sine  of  45°  and  the  cosine  of  45°  are  each  equal  to  V2/2.  Check 
by  the  table,  p.  155. 

*  25.  Turning  the  figure  of  Ex.  23 
into  the  new  position  shown  below, 
find  the  sine,  the  cosine,  and  the  tangent 
of  30°.     Check  by  the  table,  p.  155. 

*  26.   Ex.  11  may  be  restated  as  fol- 
lows ;  if  the  tangent  (§  177)  of  an  angle 
is  3,  what  is  the  cosine  of  that  angle? 
the  cosine  of  the  angle  is  1/VlO. 


Show,  by  Ex.  11,  that 
Check  by  the  table,  p.  155. 


IV,  §  202] 


MISCELLANEOUS  EXERCISES 


183 


O 


27.  How  much  leather  is  required  to  cover  a  window  seat, 
in  the  shape  of  half  of  a  regular  hexagon,  if  the  length  of  each 
of  its  three  sides  is  5  ft.  ? 

28.  Show  how  to  construct  a  line  parallel  to  the  bases  of  a 
given  parallelogram  or  triangle  that  shall  divide  the  figure 
intotwo  parts  that  are  equal  in  area. 

^9.    Prove  the  Pythagorean  theorem  by  means  of  some  one 
of  the  adjoining  figures. 

C  B        G  ^G 

K 

E 

H 
J        r  L 

(a) 

Note.  Figure  (a)  is  said  to  be  the  one  used  by  Pythagoras, 
who  discovered  the  theorem. 

Figure   (6)   is  due   to  Bhaskara,  a  native  of   India,  about 

1150  A.D. 

The  figure  used  in  §  196  is  that  used  by  Euclid,  who  wrote  a 
famous  treatise  on  Geometry  about  250-300  b.c. 

30.  Prove  the  fact  stated  in  Ex.  9,  p.  167,  by  means  of  §  189. 

31.  A  room  is  18  ft.  long,  14  ft.  wide,  and  10  ft.  high. 
Find  the  length  of  one  diagonal  of  the  floor.  Then  find  the 
length  of  a  string  stretched  through  the  center  of  the  room 
from  one  corner  at  the  floor  to  the  farthest  opposite  corner  at 
the  ceiling.  Ans.  24.9  ft. 

32.  If  n  is  any  integer,  show  that  2  w,  n^—  1,  and  n  -|- 1  are 
integers  that  are  proportional  to  the  three  sides  of  a  certain 
right  triangle.  Find  these  three  integers  when  w  =  2;  when 
w  =  4 


184 


AREAS 


[IV,  §  202 


33.  Any  polygon  described  on  the 
hypotenuse  of  any  right  triangle  as 
one  side  is  equal  to  the  sum  of  the 
two  similar  polygons  drawn  on  the 
sides,  respectively,  with  the  corre- 
sponding side  as  one  side  of  the 
polygon.     Prove  this  statement. 

[Hint.  Suppose  the  similar  polygons 
are  triangles,  as  in  the  figure.    Then 

P/^  =  a7c2,  B/q  =  byc^ ;  hence  Q  =  P -^  B.] 

^34.    Show  how  to  construct  the  side  of  an  equilateral  triangle 
whose  area  is  the  sum  of  two  given  equilateral  triangles. 

35.  In  any  triangle  ABC  let  us  write 
s  =  (a  4-  &  +  c)/2,  where  a,  b,  c,  are  the 
three  sides.     Show^  that 


Area  A  ABC  ==  V s(.s  —  a){s—  b){s  —  c). 

Outline  of  Proof.     Let  A  be  an  acute 
angle,  and  let  h  be  the  altitude  from  (7;  then  we  have 

h'  =  a'-AD\ 

b'  =  a'-{-c'-2c'AD,   by  §199; 


but 
hence 


h' 


2  r«' 


+  C2  -  b^' 


2c 


by 


But 


4c2 

=  [2  ac-\-(a^-\-c'  -  b')'][2  ac-(a'  +  c' -  b')-] -^ ^  c". 

2ac  +  (a'-\-c^-  b')  =  (a^  +  2  ac  +  c^)  -  b' 

=  (a  +  cf-b^={a-{-c-{-b){a  +  c-b) 

.  /a  -j-  &  +  c\  /a  4-  &  +  c      ,  \      .     /       ,  n 
=  %— 2— "j  (2 b^isis-  b). 

Likewise 

2ac-(a^-^c'-b')=b'-(a-cy  =  4.(s-c)(s-a), 
Hence  /i^  =  4  s  (s  —  a) (s  —  b)(s  —  c)  -h  c^, 

and  Area  A  ABC  =  h  •  c/2  =  Vs(s  —  a)(s  —  b){s  —  c), 


IV,  §202]  MISCELLANEOUS  EXERCISES  185 

36.  Find  the  area  of  a  triangular  field  whose  sides  are,  re- 
spectively, 80  rd.,  220  rd.,  200  rd.     [1  acre  =  160  sq.  rd.] 

37.  Find  the  area  of  a  parallelogram  whose  sides  are,  re- 
spectively, 8  in.  and  10  in.  long,  and  one  of  whose  diagonals  is 
15  in.  long.  Ans.   74.0  sq.  in. 

38.  Find  the  area  of  a  triangle,  given  two  sides  15  ft.  and 
25  ft.,  and  their  included  angle  30°. 

*  39.  Find  the  area  of  a  triangle,  given  two  sides  15  ft.  and 
25  ft.,  and  their  included  angle  40°.  Ans.  120.5  ft. 

40.  Show  that  the  area  of  any  triangle  is  s  •  ?•,  where 
s  =  (a-{-b+c)/2,  as  in  Ex.  35,  and  r  is  the  radius  of  the  in- 
scribed circle.     (See  §  124.) 

41.  Show  that  the  area  of  any  polygon  circumscribed  about 
a  circle  is  half  the  sum  of  its  sides  times  the  radius  of  the 
circle. 

42.  Comparing  the  two  results  of  Exs.  35  and  40,  show  that 
the  radius  of  a  circle  inscribed  in  any  triangle  is 

r  =  V(s  —  a)(s  —  b){s  —  c)/s. 

43.  Show  that  in  any  triangle  one  of  whose  angles  is  120°, 
the  square  on  the  side  opposite  the  larger  angle  equals  the  sum 
of  the  squares  on  the  other  two  sides  plus  the  product  of  those 
sides. 

44.  Show  that  in  any  triangle  one  of  whose  angles  is  60°  the 
square  on  the  side  opposite  that  angle  is  equal  to  the  sum  of 
the  squares  on  the  other  two  sides  diminished  by  the  product 
of  these  two  sides. 

*  45.  Show,  in  Figs.  138  and  139,  that  h  =  bx  (sine  of  Z  C): 
hence  show  that  the  area  of  the  triangle  ABC  is 

Area  A  ABC=\  ah  X  (sine  of  Z  C). 

*  46.   Prove  Theorem  IV,  §  193,  by  means  of  Ex.  45. 


CHAPTER  V 

REGULAR   POLYGONS   AND   CIRCLES 

203.  Regular  Polygon.  A  polygon  that  is  both  equiangular 
and  equilateral  is  called  a  regular  polygon.     (See  §  96.) 

EXERCISES 

1.  Construct  a  quadrilateral  that  is  equilateral  but  not 
equiangular.     What  are  such  quadrilaterals  called  ? 

2.  Construct  a  quadrilateral  that  is  equiangular  but  not 
equilateral.     What  are  such  quadrilaterals  called  ? 

3.  Draw  a  quadrilateral  that  is  neither  equiangular  nor 
equilateral. 

4.  Construct  a  regular  polygon  of  four  sides.  What  are 
such  quadrilaterals  called  ? 

5.  Is  an  equilateral  triangle  a  regular  polygon  ?     Why  ? 

6.  What  is  the  size  of  each  angle  of  a  regular  polygon  of 
seven  sides  ?     (See  §  97.) 

7.  If  three  equal  rods  are  hinged  together  at  their  ends  in 
the  form  of  an  equilateral  triangle,  is  the  framework  rigid  ? 

8.  If  four  equal  rods  are  joined  together  at  their  ends  by 
hinges,  is  the  framework  formed  necessarily  a  square  ?  Can 
the  angles  be  changed  without  changing  the  lengths  of  the  sides  ? 

9.  If  five  equal  rods  are  hinged  together  at  their  ends  in 
the  form  of  a  regular  pentagon,  is  the  framework  rigid  ? 

10.  Will  the  framework  of  Ex.  8  be  rigid  if  a  stiff  rod  is  in- 
serted along  one  diagonal  ?  Along  how  many  diagonals  must 
rods  be  inserted  in  the  framework  of  Ex.  9  to  make  it  rigid  ? 

186 


V,  §204]  MUTUAL  RELATIONS  187 

204.   Theorem  I.    If  a  circle  is  divided  into  a  number 
of  equal  arcs : 

(a)  the  chords  joining  the  points  of  division  form  a 
regular  inscribed  polygon  ; 

(b)  tangents  draivn  at  the  points  of  division  form,  a 
regular  circumscribed  polygon. 

s 


Given  the  circle  divided  into  the  equal  arcs  AB,  BC,  CD, 
DE,  and  EA  by  the  chords  AB,  BC,  CD,  DE,  and  EA ;  and 
given  the  tangents  PQ,  QR,  RS,  ST,  and  TP,  touching  the 
circumference  at  the  points  B,  C,  D,  E,  and  A,  respectively. 

To  prove  that  (a)  ABCDE  is  a  regular  polygon ; 
(h)  PQRST  is  a  regular  polygon. 

Proof  of  (a).     Since  AB=BC=  CD=DE=EI),  Given 

Arcs  ABC,  BCD,  CDE,  DEA,  and  EAB  are  equal.     Why  ? 
Hence  Z  ABC  =  Z  BCD  =  Z  CDE,  etc.  Why  ? 

Also  AB  =  BC=CD=  DE,  etc.  Why  ? 

It  follows  that  ABCDE  is  a  regular  polygon.  §  203 

Proof  of  (6).      Show  that  A  APB  ^  A  BQC  by  showing  that 

AB  =  BC,  Z  PAB  =  Z  QBC  (§  137),  and  Z  PBA  =  Z  QCB. 

Likewise  A  APB  ^  A  CRD,  etc.     Then  show  that  PQ  =  QR 

=  etc. ;  and  Z  P  =Z  Q  =  etc. ;  and   thus   prove   the   polygon 

PQRST  a  regular  polygon. 


188 


REGULAR  POLYGONS  AND  CIRCLES    [V,  §  205 


205.  Theorem  II.    (a)  A  circle  may  he  circumscribed 
about  any  regular  polygon ; 

(b)  a  circle  may  also  be  inscribed  in  it. 


Given  the  regular  polygon  ABCDE. 

To  prove  that  (a)  a  circle  may  be  circumscribed  about  it ; 

(p)  Si  circle  may  be  inscribed  in  it. 
Proof  of  (a).     Pass  a  circle  through  the  points  A,  Bj  and  C. 

§120 
Join  the  center  0  with  the  vertices  of  the  polygon. 


Then 

Moreover 
and 

Therefore 
so  that 


CD. 


§§  103,  203 

§  203 

§40 

Ax.  2 

§35 

Why? 


OB  =  OC,  and  AB 
Z  CBA  =  Z  DCB, 
Z  CBO  =  Z  OCB. 
Z  OB  A  =  Z  OCD, 
AOAB^A  OCD', 
hence  OA  =  OD,  and  the  circle  passes  through  D. 

In  like  manner  show  that  the  circle  passes  through  E. 

Hence  the  circle  is  circumscribed  about  the  polygon.     §  114 

Proof  of  (&).     The  sides  of  the  regular  polygon  are  equal 

chords  of  the  circumscribed  circle.    Therefore  they  are  equally 

distant  from  the  center.  §  109 

The  circle  with  0  as  a  center  and  the  perpendicular  from  the 

center  to  any  side  as  radius  is  inscribed  in  the  polygon.    §  114 


V,  §207]  MUTUAL  RELATIONS  189 

206.  Definitions.  The  common  center  of  the  circumscribed 
and  inscribed  circles  is  called  the  center  of  the  regular  polygon. 

The  radius  R  of  the  circumscribed  circle  is  called  the  radius 
of  the  regular  polygon. 

^ ^D 


The  radius  r  of  the  inscribed  circle  is  called  the  apothem  of 
the  regular  polygon. 

The  perimeter  of  a  polygon  is  the  sum  of  the  lengths  of  its 
sides. 

207.  Theorem  III.  The  area  of  a  regular  polygon  is 
equal  to  half  the  product  of  its  apothem  and  perimeter. 

Given  the  regular  polygon  ABODE  ••♦,  Fig.  145.  Let  its 
perimeter  be  denoted  by  p  and  its  apothem  by  r. 

To  prove  that  ABCD  •-.  =pr/2. 

Proof.  Draw  the  radii  OA,  OB,  0(7,  etc.,  Fig.  145,  thus 
making  as  many  triangles  as  the  polygon  has  sides. 

The  altitude  of  each  triangle  is  the  apothem  r,  and  each  base 
is  a  side  of  the  polygon ;  hence,  the  area  of  each  triangle  is 
half  the  product  of  r  and  one  side  of  the  polygon.  §  189 

Since  the  sum  of  all  the  bases  is  the  perimeter  p  of  the  poly- 
gon, and  the  sum  of  the  areas  of  the  triangles  is  the  area  of 
the  polygon,  it  follows  that 

Area  of  ABCD  •••  =pr/2. 

Compare  this  result  with  Ex.  41,  p.  185. 


190  REGULAR  POLYGONS  AND  CIRCLES    [V,  §  207 

EXERCISES 

1.  If  n  represents  the  number  of  sides  of  a  regular  polygon, 
show  that  the  angle  at  the  center  subtended  by  one  side  is  4/ii 
right  angles. 

2.  Find  the  angle  at  the  center  subtended  by  one  side  of  an 
equilateral  triangle ;  of  a  square ;  of  a  regular  pentagon  ;  of  a 
regular  hexagon.    Draw  these  figures  by  means  of  a  protractor. 

3.  Prove  that  any  angle  of  a  regular  polygon  is  supple- 
mentary to  the  angle  at  the  center  subtended  by  one  side. 

4.  Eind  the  number  of  degrees  in  the  angle  at  the  center 
subtended  by  one  side  of  a  regular  octagon.  Eind  the  size  of 
one  angle  of  the  octagon  by  means  of  Ex.  3 ;  and  then  check 
your  answer  by  §  97.     Draw  the  figure. 

5.  How  many  sides  has  a  regular  polygon  if  the  angle  at 
the  center  subtended  by  one  side  is  40°  ? 

6.  If  one  angle  of  a  regular  polygon  is  140°,  what  is  the 
number  of  its  sides  ? 

7.  Eind  the  area  of  a  regular  hexagon  whose  side  is  6  in. 

8.  Two  squares  have  for  their  sides  8  and  11  in.,  respec- 
tively.    Eind  by  §  207  the  ratio  of  their  areas. 

9.  Two  squares  have  areas  of  144  sq.  in.  and  225  sq.  in., 
respectively.     What  is  the  ratio  of  their  perimeters  ? 

^10.   Eind  the  ratio  of  the  perimeters  of  squares  inscribed  in 
and  circumscribed  about  the  same  circle.  Ans.  1 :  V2. 

11.  The  perimeter  of  a  regular  inscribed  hexagon  is  30  in. 
Find  the  perimeter  of  a  regular  hexagon  circumscribed  about 
a  circle  of  twice  the  diameter.  Ans.  69.28  in. 

12.  Eind  the  area  of  an  equilateral  triangle  circumscribed 
about  a  circle  whose  radius  is  12  in.  Ans.  748.25  sq.  in. 

13.  Find  the  apothem  of  an  equilateral  triangle  of  which  one 
side  is  6  in.  (Use  Th.  XXXIII,  §  102.)  Hence  find  its  area 
by  Theorem  III.     Compare  with  the  result  of  Ex.  6,  p.  174. 


V,  §  210] 


MUTUAL  RELATIONS 


191 


208.  Areas  of  Inscribed  and  Circumscribed  Regular  Poly- 
gons. The  area  of  a  circle  is  evidently  greater  than  the  area 
of  any  regular  polygon  inscribed  in 
it.     (17,  §31.) 

By  doubling  the  number  of  sides, 
we  obtain  regular  inscribed  poly- 
gons whose  areas  are  more  nearly 
equal  to  that  of  the  circle,  as  in- 
dicated in  the  figure. 

Thus  the  area  of  a  regular  in- 
scribed octagon,  for  example,  is  evi- 
dently more  nearly  equal  to  the 
area  of  the  circle  than  is  the  area 
of  an  inscribed  square. 

The  area  of  a  circle  is  also  evidently  less  than  the  area  of 
any  circumscribed  polygon. 


Perimeters.  The  perimeter  of  any  inscribed  polygon 
is  evidently  less  than  the  length  of  the  circumference  (Post.  3, 
§  28) ;  we  shall  also  assume  that  the  length  of  the  circumfer- 
ence is  less  than  that  of  any  circumscribed  polygon. 

210.  Areas  and  Lengths  of  Circles.  Since  the  regular 
inscribed  and  regular  circumscribed  polygons  approach,  both  in 
length  and  in  area,  nearer  to  one  another  and  to  the  circle, 
as  the  number  of  sides  is  doubled,  we  may  say : 

If  the  number  of  sides  of  the  regular  inscribed  and  regular 
circumscribed  polygons  is  repeatedly  doubled^ 

(a)  their  areas  approach  the  area  of  the  circle  as  a  common 
limit; 

(b)  their  perimeters  approach  the  length  of  the  circumference 
of  the  circle  as  a  common  limit. 

It  is  also  obvious  that  the  apothem  of  the  inscribed  regular 
polygon  will  approach  the  radius  of  the  circle,  under  the  same 
circumstances. 


192  REGULAR  POLYGONS  AND   CIRCLES    [V,  §211 

211.   Theorem  IV.     The  circumferences  of  two  circles  are  to 
each  other  as  their  radii. 


Fig.  147 

Given  two  circles  0  and  0'  whose  radii  are  r  and  r'. 

To  prove  that  the  length  of  their  circumferences  c  and  c'  are 
to  each  other  as  their  radii ;  that  is, 

c/c^  =  T/r\ 

Proof.  Draw  inscribed  regular  polygons  P  and  P\  of  the 
same  number  of  sides  in  each  circle,  let  the  perimeters  of  these 
polygons  be  p  and  p',  and  let  the  corresponding  sides  be  s  and  s'. 

Then  P^P',  §166 

and  p/p'  =  s/s'.  §  167 

But  s/s'  =  r/r', 

since  the  triangle  two  of  whose  sides  are  s  and  r  is  similar  to 
the  triangle  two  of  whose  sides  are  s'  and  r'. 

Hence  p/p'  =  r/r'.  Ax.  9 

Since  p  and  p'  approach  c  and  c'  if  the  number  of  sides  is 
repeatedly  doubled,  the  difference  p/p'  —  c/c'  can  be  made  as 
small  as  we  please  by  doubling  the  number  of  sides  repeatedly; 
hence  also  r/r'  —  c/c'  can  be  made  as  small  as  we  please. 

But  r/r'  and  c/c'  do  not  change  at  all  as  we  increase  the 
number  of  sides  of  the  polygons ;  hence 

r/r'  =  c/c', 
for  if  r/r'  and  c/c'  differed  from  each  other,  their  difference 
would  be  fixed,  and  therefore  not  as  small  as  we  might  please. 


V,  §214]          LENGTH  OF  CIRCUMFERENCE  193 

212.  Corollary  1.  The  ratio  of  a  circumference  to  its  diameter 
is  the  same  for  all  circles. 

For,  since  c/c'  =  r/r'  (§  211),  we  have  also,  if  d  and  d'  are  the 
diameters,  d/d'  =  2  r/2  r'  =  r/r'  =  c/c'. 

Hence  by  alternation  (Theorem  D,  §  144),  c/d  =  c'/d'. 

213.  The  Number  ir.  The  number  obtained  by  dividing  the 
circumference  of  any  circle  by  its  diameter,  which,  by  §  212,  is  the 
same  for  all  circles,  is  denoted  by  the  Greek  letter  tt  (pronounced 
pi).  This  number,  which  is  about  3^  (or,  more  accurately  still, 
3.1416),  will  be  computed  approximately  later  in  this  chapter. 

214.  Corollary  2.  In  any  circle  c  =  ird,  or  c  =  2  ttt,  ivhere  r 
is  the  radius,  d  the  diameter,  and  c  the  length  of  the  circumference. 

For,  since  tt  =  -  (§  213),  c  =  ird]  or  since  d=2  r,  c  =  2 Trr. 
d 

EXERCISES 

1.  Find  the  length  of  the  circumference  of  a  circle  of  radius 
2  ft.     [Take  tt  =^  3f  ]  Ans.  4  n  ft.,  or  12.57  ft. 

2.  If  the  radius  B  of  the  earth  is  4000  mi.,  what  is  its 
circumference  at  the  equator?  [Take  tt  =  3}.]  How  much 
is  this  result  affected  by  taking  the  more  accurate  values 
E  =  3963  mi.,  tt  =  3.1416. 

3.  Find  the  diameter  of  a  circle  whose  circumference  is  40  in. 

4.  Measure  the  circumference  of  some  round  object,  such 
as  a  porch  column,  by  stretching  a  string  around  it  tightl}^  and 
then  measuring  the  string.  '  Now  compute  the  diameter. 

5.  How  wide  must  a  piece  of  tin  be  cut  in  order  to  be  made 
into  a  stovepipe  8  in.  in  diameter  ?  Ans.  8  n  in.,  or  25f  in. 

6.  Show  how  to  find  quickly  the  approximate  length  of 
wire  in  a  coil  by  measuring  the  diameter  of  the  coil,  and  count- 
ing the  number  of  strands. 

7.  What  is  the  length  of  an  arc  that  subtends  an  angle  of 
60°  at  the  center  of  a  circle  whose  radius  is  5  ft.  ? 


194 


REGULAR  POLYGONS  AND  CIRCLES    [V,§21o 


215.  Theorem  V.     The  area  of  a  circle  is  equal  to  one 
half  the  product  of  its  radius  and  its  circumference. 


Fig.  148 

Given  the  circle  O,  with  radius  r,  circumference  c,  and 
area  A. 

To  prove  that  A  =  rc/2. 

Proof.  Circumscribe  a  regular  polygon  about  the  circle. 
Let  A  denote  its  area  andy  its  perimeter. 

Then  A'  =  rp' /2.  §  207 

As  the  number  of  sides  of  the  regular  circumscribed  polygon 
is  increased,  p'  approaches  c  as  its  limit.  §  210 

Hence  rp' /2  approaches  rc/2  as  a  limit. 

Also  A'  approaches  ^  as  a  limit.  §  210 

Therefore  the  difference  between  A  and  rc/2  must  be  as 
small  as  we  please,  as  in  §  211.  It  follows,  as  in  §  211,  that 
A  =  rc/2. 

The  student  should  state  carefully  all  of  the  remaining  steps 
in  the  argument. 

216.  Corollary  1.  The  area  of  a  circle  is  equal  to  tt  times  the 
square  of  its  radius,  that  is,  A  =  irr^. 

For,  by  §  215,  A  =  rc/2 ;  but  c  =  2  Trr ;  hence  A  =  ttt^. 

217.  Corollary  2.  The  areas  of  two  circles  are  to  each  other 
as  the  squares  of  their  radii. 


V,  §  219]  AREA  OF  CIRCLE  195 

Note.  The  very  famous  problem  of  "  squaring  the  circle," 
that  is,  of  constructing  the  side  of  a  square  whose  area  equals 
that  of  a  given  circle,  depends  on  determining  the  value  of  tt. 
We  now  know  that  this  construction  is  impossible  with  ruler 
and  compasses ;  but  the  ancient  Greeks  and  the  Schoolmen  of 
the  Middle  Ages  spent  much  time  attempting  to  do  it. 

218.  Sectors.  The  area  of  a  sector  bears  the  same  ratio 
to  the  area  of  a  circle  as  the  angle  of  the  sector  bears  to  360°. 
For  example,  the  area  of  a  sector  whose  arc  is  36°  is  1/10  the 
area  of  the  circle. 

219.  Circle  divided  into  Sectors.  A  circle  may  be  cut  into 
equal  sectors  and  arranged  as  in  the  following  figure. 


Fig.  149.    Circle  divided  into  Sectors 

Each  sector  approximates  the  shape  of  a  triangle.  The  area 
of  each  sector  is  equal  to  one  half  the  product  of  its  arc  (base) 
by  its  radius  (altitude).  Their  sum  equals  one  half  the  whole 
circumference  times  the  radius. 

EXERCISES 

[Exercises  that  involve  the  number  v  should  be  worked  through  first 
in  terms  of  the  symbol  v  ;  then  the  value  2>\  should  be  substituted  for  it. 
Thus  the  area  of  ^  of  a  circle  whose  radius  is  6  inches  is  ^  •  6^ .  tt  sq.  in. 
=  9  TT  sq.  in.  ;  substituting  t  =  3f,  we  find  the  result  28f  sq.  in.] 

1.  Find  the  area  of  a  circle  whose  diameter  is  14  in. 

2.  Find  the  area  of  a  square  circumscribed  about  a  circle 
whose  radius  is  2  ft.  What  is  the  ratio  of  the  area  of  the 
circle  to  the  area  of  this  square  ?    Ans.  16  sq.  ft. ;  7r/4,  or  11/14. 


196  REGULAR  POLYGONS  AND  CIRCLES    [V,§219 

3.  The  circumference  of  a  circle  is  14  ft.  Find  its  radius 
and  its  area. 

4.  The  area  of  a  circle  is  24  sq.  in.     Find  its  radius. 

5.  What  is  the  change  in  the  area  of  a  circle,  if  its  radius 
is  multiplied  by  2  ?  by  5  ?  loj  n  ? 

6.  The  effect  of  an  explosion  in  a  large  powder  plant  was 
felt  for  a  distance  of  10  mi.  in  every  direction.  How  large  an 
area  was  affected  ? 

7.  A  tinner  is  to  cut  the  largest  possible  square  from  a  cir- 
cular piece  of  tin.     What  proportion  of  the  tin  will  be  left  ? 

8.  Find  the  area  of  a  sector  in  a  circle  of  radius  3  ft.  if  the 
angle  of  the  sector  is  45°  ;  60° ;  30°  ;  80°. 

9.  If  a  tinner  cuts  from  the  same  sheet  of  tin  two  circular 
pieces,  one  of  which  is  twice  as  wide  as  the  other,  how  much 
heavier  is  the  larger  one  ? 

10.  If  it  is  desired  to  cut  two  circular  weights  out  of  a  flat 
piece  of  metal,  how  much  wider  must  the  larger  be  to  weigh 
twice  as  much,  the  thickness  being  the  same  ? 

11.  A  steel  rod  whose  cross-section  is  1  in.  square  weighs 
3.4  lb.  per  foot  of  length.  Find  the  weight  per  foot  of  length 
of  a  round  rod  2  in.  thick  of  the  same  material. 

12.  Find  the  weight  per  foot  of  length  of  a  waterpipe  whose 
outer  diameter  is  3  in.,  and  whose  inner  diameter  is  2.75  in., 
made  of  the  material  mentioned  in  Ex.  11. 

13.  Show  that  the  area  of  a  circular  ring      /^^^^^^>\ 
contained  between  two  concentric  circles  of     '^W    t^-"^^^ 
diameters  D  and  d  is  ^^    ^'^^^ 

B-d   D  +  d  '\^^^^W\ 


(UP-  -d'\ 


A=7ri  : 1,  Or^  =  7r 


r<-t-*k- — d— >t*i-»; 


14.   How  much  water  per  hour  (in  cubic 
feet)  will  flow  through  a  pipe  whose  inner  diameter  is  3  in.. 
if  the  water  is  flowing  3  ft.  per  second  ? 


V,  §  221] 


AREA  OF  CIRCLE 


197 


220.  Problem  1.  Given  the  side  and  radius  of  a  regular  in- 
scribed polygon,  to  find  the  side  of  a  regular  inscribed  polygon  of 
double  the  number  of  sides. 


Given  AB,  the  side  of  a  regular  inscribed  polygon  of  radius  r. 

Required  to  find  AC,  a.  side  of  the  regular  inscribed  polygon 
of  double  the  number  of  sides. 

Solution.  Draw  the  diameter  CEj  and  the  radius  AO.  Also 
draw  AE. 

Now  OD  ±  AB  at  its  middle  point. 


Hence 
or 
and 

Also 
that  is, 


Olf^r'-iAB", 


OD  =  Vr^  J  J^, 


CD  =  r-  vy  -  \  aW. 
AC^=CE'CD  =  2r'CD; 

AC=^2r(r--Vr^-iAF 

=  Vr(2  r  -  V4  r^  -  AW). 


Why? 
Why? 


163 


221.    Corollary  1.     If  r  =  1,  and  s  =  the  side  of  the  inscribed 
polygon,  


198 


REGULAR  POLYGONS  AND  CIRCLES    [V,  §222 


222.    Problem  2.     To  compute  approximately  the  value  of  tt. 

The  perimeter  of  a  regular  hexagon  inscribed  in  a  circle  of 
unit  radius  is  6  units.  (Why  ?)  By  using  the  formula  in  §  221 
and  computing  successively  the  perimeters  for  polygons  of  12, 
24,  52,  .  •  •,  and  768  sides  we  get  the  following  results : 


NiTMBEK  OF  Sides 

Length  op  one  Side 

Length  of  Pekimeteb 

12 

.51763809 

6.21165708 

24 

.26105238 

6.26525722 

48 

.13080626 

6.27870041 

96 

.06643817 

6.28206396 

192 

.03272346 

6.28290510 

384 

.01636228 

6.28311544 

768 

.00818126 

6.28316941 

By  continuing  this  process  it  is  found  that  the  first  five 
figures  in  the  decimal  remain  unchanged.  Hence  6.28317  is 
a  close  approximation  to  the  circumference  of  a  circle  whose 
radius  is  1.  Since  the  diameter  is  2,  the  ratio  tt  of  the  circum- 
ference to  the  diameter  is,  approximately, 

IT  =  5:?^  =  3.14159  (usually  written  3.1416,  or  3|). 


Note.  A  still  more  accurate  value  can  be  computed  by  con- 
tinuing the  preceding  process.  It  has  been  proved  that  the 
number  tt  cannot  be  expressed  precisely  by  any  finite  decimal. 

The  fact  that  tt  cannot  be  expressed  precisely  is  equivalent 
to  the  statement  that  the  diameter  and  the  circumference  of  a 
circle  are  incommensurable  to  each  other  (§  128).  But  we  can 
obtain,  by  the  preceding  process,  as  great  accuracy  as  we  please. 

The  value  is  known  to  over  700  decimal  places.  To  ten 
places  it  is  TT  =  3.1415926536,  but  such  accuracy  is  never  neces- 
sary in  any  ordinary  affairs.    As  a  curiosity,  we  quote  the  value : 

TT  =  3.1415926535897932384626433832795028841971693993751. 


V,  §  224] 


AREA  OF  CIRCLE 


199 


223.  Problem  3.     To  inscribe  a  square 
in  a  given  circle. 

Given  the  circle  0. 

To  inscribe  a  square  in  circle  0.  ^ 

Construction.     Draw  two  diameters  AC 
and  DB  perpendicular  to  each  other. 
Draw  AB,  BC,  CD,  and  DA. 
Then  ABCD  is  the  square  desired. 
Proof.     [The  proof  is  left  for  the  student.] 

224.  Problem  4.  To  inscribe  a  regular  hexagon  in  a  given 
circle.  I  g 

Given  the  circle  0. 

To  inscribe  a  regular  hexagon  in  circle  0. 

Construction.  Draw  the  radius  OA  and 
with  ^  as  a  center  and  radius  OA  draw  an 
arc  cutting  the  circle  in  B.  Then  AB  is 
the  side  of  the  hexagon  desired. 

Outline  of  Proof 
central  angle  60°. 

EXERCISES 

1.  Show  how  to  inscribe  an  equilateral  triangle. 

2.  Show  how  to  inscribe  a  regular  polygon  of  twelve  sides. 

3.  Since  one  side  of  an  inscribed  square  subtends  a  central 
angle  90°,  and  one  side  of  a  regular  inscribed  hexagon  subtends 
a  central  angle  of  60°,  show  that  if  one  vertex  of  the  square 
coincides  with  one  vertex  of  the  hexagon,  the  next  vertices  are 
at  the  extremities  of  an  arc  of  30°. 

4.  Show  how  to  inscribe  a  regular  polygon  of  eight  sides. 
^6.    Show  how  to  inscribe  a  regular  polygon  of  twenty-four 
sides  directly  from  an  inscribed  regular  octagon  and  a  regular 
inscribed  hexagon. 


Fig.  152 
Draw  OB  and  show  that  AB  subtends  a 


200 


REGULAR  POLYGONS  AND  CIRCLES    [V,  §225 


225.    Problem  5.     To  inscribe  a  regular  decagon. 

Given  the  circle  0. 

Required  to  inscribe  a  regular  decagon  in  the  given  circle. 


Fig.  153 


Construction.  Draw  the  radius  OA  and  divide  it  in  extreme 
and  mean  ratio  at  O,  having  the  larger  segment  next  to  the 
center.  §  172 

Then  OA:OG  =  OG:  GA,  and  OG  is  the  side  of  the  decagon 
required.  By  applying  OG  ten  times  to  the  circle  as  a  chord, 
the  desired  regular  decagon  is  formed. 

Proof.     Draw  OB  and  BG. 

Since  OA/OG  =  OG/GA  and  OG  =  AB, 

we  have  OA/AB  =  AB/  GA ; 

hence  A  OAB  ~  A  ABG. 

Therefore  ZO  =  Z  ABG  and  Z  AGB  =  Z  ABO. 

But  Z  OAB=  ZABO', 

hence  Z  AGB  =  Z  GAB,  and  AB  =  BG  =  GO. 

Therefore  Z  0  =  Z  GBO. 

But  we  have  shown    Z  0  =  Z  ABG ; 
hence,  addin'g,       2  ZO=Z  OBA  =  Z  OAB. 

Now  ZO-\-Z  OBA  +  Z  OAB  =  2  rt.  A, 

whence  5  Z  0  =  2  rt.  z^, 

or  Z0  =  1/5  of  2  rt.  A,  or  1/10  of  4  rt.  A. 

Therefore  the  arc  AB  is  1/10  of  the  circumference,  and  the 
chord  AB,  equal  to  OG,  is  a  side  of  a  regular  inscribed  decagon. 


Const. 
Why? 
Why? 
Why? 
Why? 
Why? 
Why? 


Why? 

Why? 


V,  §  225]  AREA  OF  CIRCLE  201 

EXERCISES 

1.   Show  how  to  inscribe  a  regular  pentagon  in  a  circle. 
2    Show  how  to  inscribe  regular  polygons  of  20,  40,  etc. 
sides  in  a  given  circle. 

3.  Show  that  if  one  vertex  of  a  regular  inscribed  pentagon 
coincides  with  one  of  a  regular  hexagon  inscribed  in  the  same 
circle,  the  uext  vertices  are  extremities  of  an  arc  of  12°.  Hence 
show  how  to  inscribe  a  regular  polygon  of  30  sides. 

4.  Show  how  to  inscribe  a  regular  polygon  of  15  sides 
directly  by  using  the  regular  inscribed  pentagon  and  the  in- 
scribed equilateral  triangle. 

MISCELLANEOUS  EXERCISES.    CHAPTER  V 

1.   The  radius  of  a  circle  is  2  in.     Find  the  length  of  the 
circumference ; .  the  area, 
w^  2.   The  area  of  a  circle  is  98  sq.  ft.     Find  the  diameter. 

3.  The  diameters  of  two  circles  are  4  ft.  and  9  ft.,  respec- 
tively.    Find  the  ratio  of  their  areas. 

4.  How  many  people  can  be  seated  at  a  round  table  54  in. 
in  diameter  when  it  is  extended  4  ft.,  allowing  2  ft.  to  a  person  ? 

5.  Find  the  perimeter  of  a  regular  hexagon  inscribed  in  a 
circle  whose  radius  is  1  ft. ;  3  f t. ;  a  ft. 

-  6.  The  perimeter  of  a  regular  inscribed  hexagon  is  48  ft. 
What  is  the  diameter  of  the  circle  ? 

7.  Find  the  perimeter  of  a  regular  circumscribed  hexagon, 
if  the  radius  of  the  circle  is  1  ft. ;  3  ft. ;  r  ft. 

8.  If  the  radius  of  a  circle  is  r,  find  the  area  of  the  in- 
scribed equilateral  triangle;  of  the  circumscribed  equilateral 
triangle. 

>  9.  Show  that  the  area  of  the  inscribed  equilateral  triangle 
equals  one  fourth  the  area  of  the  circumscribed  equilateral 
triangle. 


202  REGULAR  POLYGONS  AND  CIRCLES    [V,  §225 

10.  A  cow  is  tethered  at  the  end  of  a  50  ft.  rope,  which  is 
fastened  to  the  corner  of  a  barn.  The  barn  is  25  ft.  wide  and 
60  ft.  long.     Over  how  much  area  may  the  cow  graze  ? 

11.  How  many  revolutions  per  mile  does  a  28-in.  bicycle 
wheel  make  ?  Ans.    720. 

12.  The  boiler  of  an  engine  has  200  tubes,  each  3  in.  in 
diameter,  for  conducting  the  heat  through  the  water.  Find 
their  total  cross  sectional  area. 

13.  Construct  a  regular  inscribed  pentagon,  and  draw  all  the 
diagonals.  Show  that  the  sum  of  all  the  angles  in  the  vertices 
of  the  resulting  five-pointed  star  equals  two  right  angles. 

14.  A  circular  piece  of  brass  has  a  radius  of  10  in.  and  it  is 
desired  to  cut  a  hole  through  it  equal  in  area  to  one  half  the 
disk.     What  should  be  the  radius  of  the  hole  ? 

15.  The  central  angle  whose  arc  is  equal  to  the  radius  of 
the  circle  is  called  a  radian.  It  is  often  used  as  a  unit  of 
measure  of  angles.  Show  that  1  radian  =  180°  -r-7r=  57.3° 
approximately. 

^16.  A  circle  is  circumscribed  about  a  right  triangle  and  two 
others  are  described  with  the  sides  as  diameters.  Prove  that 
the  large  circle  equals  the  sum  of  the  two 
small  ones.  r    ^mmSmm^)^ 

17.  Semicircles  are  constructed  on  the 
three  sides  of  a  right  triangle  as  in  the 
adjacent  figure.  Show  that  the  sum  of 
the  semicircles  AO'CD  and  BO"CE  is 
equal  to  the  semicircle  AOBF.  ^'^mMB^Q 

18.  Construct  a  circle  equal  to  the  area  of  two  given  circles. 

v^  19.  If  the  limit  of  safety  for  the  surface  speed  of  an  emery 
stone  is  5500  ft.  per  minute,  what  is  the  diameter  of  the 
largest  wheel  that  can  safely  make  1500  revolutions  per 
minute?  Ans.  11/(3  tt)  ft.,  or  14  in. 


V,  §225]  MISCELLANEOUS  EXERCISES  203 

20.  Can  a  piece  of  paper  6  in.  wide  be  used  to  wrap  up  a 
circular  mailing  roll  whose  radius  is  1  in.  ? 

21.  How  many  laps  around  a  circular  running  track  whose 
diameter  is  125  yd.  are  necessary  to  make  up  a  distance  of 
8  mi.? 

22.  Find  the  length  of  the  curved  portion  of  a  railroad  track 
that  connects  two  straight  portions  at  right  angles  to  each 
other,  given  that  the  curved  portion  is  an  arc  of  a  circle  of 
radius  200  ft.,  tangent  at  its  extremities  to  the  straight  portions 
of  the  track. 

Find  the  length  of  the  curved  portion  if  the  angle  between 
the  straight  portions  is  60°. 

23.  What  is  the  total  pressure  on  the  piston  of  an  engine 
if  the  cylinder  is  20  in.  in  diameter  and  the  gauge  shows  76  lb. 
per  square  inch  ? 

24.  Four  pumps  each  with  a  diameter  of  5  in.  are  used  in  a 
mine.  If  one  pump  were  used  to  remove  the  same  amount  of 
water  in  the  same  time,  what  would  be  its  diameter,  every- 
thing else  being  the  same?  Ans.  10  in. 

25.  When  the  gauge  shows  a  steam  pressure  of  100  lb.  per 
square  inch,  what  is  the  total  pressure  tending  to  blow  the 
cylinder  head  out,  if  it  is  18  in.  in  inside  diameter  ? 

26.  If  the  water  in  a  3-in.  (inside  diameter)  water  main  is 
flowing  at  the  rate  of  5  ft.  per  second,  how  much  water  is 
passing  a  given  point  per  minute?     1  gal.  =  231  cu.  in. 

27.  A  water  main  of  6-in.  diameter  is  continued  beyond  a 
certain  point  by  a  pipe  of  4-in.  diameter.     If  the  water  in  the 


6-in.  pipe  is  running  at  the  rate  of  2  ft.  per  second,  how  fast 
is  the  water  in  the  4-in.  pipe  running? 


204  REGULAR  POLYGONS  AND  CIRCLES    [V,  §  225 

28.  A  conduit  for  carrying  water  is  circular  in  form  and  is 
10  ft.  in  diameter.  Find  the  area  of  the  total  cross  section.  If 
the  water  level  is  at  EF,  find  the  area  of 
the  cross  section  of  the  water  if  the 
angle  EOF  is  90° ;  if  EOF  is  60°.  ^_  water 

29.  Find  the  amount  of  water  flow- 
ing through  the  conduit  of  Ex.  28  per 
minute  if  Z.EOF=  60°  and  the  speed  is 

2  ft.  per  second. 

30.  Find  the  length  EGHF  of  the 
portion  of  the  circular  outline,  Ex.  28,  which  is  wet  when  the 
water  reaches  EF  if  EOF  is  45°;  if  EOF  is  60°;  if  EOF  is 
equal  to  90°.  (This  so-called  "  wetted  perimeter "  is  of  the 
greatest  importance  in  determining  friction  and  therefore  the 
resistance  of  the  pipe  to  the  water  flow.) 

*  31.  Find  the  area  of  the  cross  section  of  the  water  in  the 
conduit  of  Ex.  28  when  Z  EOF=  80°.  Ans.  73.41  sq.  ft. 

*  32.  Find  the  area  of  the  cross  section  of  the  water  in 
the  conduit  of  Ex.  28  when  the  distance  from  0  to  EF  is 
4  ft. 

33.  How  fast  is  the  point  on  the  rim  of  a  wheel  moving,  in 
feet  per  second,  if  the  wheel  is  2  ft.  in  diameter  and  is  rotating 
at  an  angular  speed  of  four  revolutions  per  minute? 

34.  How  far  does  a  carriage  move  when  one  of  its  wheels 
revolves  (without  slipping)  through  five  complete  revolutions, 
if  the  diameter  of  the  wheel  is  4  ft.  ? 

35.  How  many  times  will  the  wheel  of  a  bicycle  revolve,  if 
it  is  28  in.  in  diameter,  in  going  3  mi.  ?     If  the  bicyclist  goes 

3  mi.  in  20  min.,  how  many  revolutions  does  the  wheel  make 
per  minute? 

36.  Find  the  angular  speed  of  a  car  wheel  that  is  20  in.  in 
diameter,  when  the  train  is  going  40  mi.  per  hour. 


V,  §  225] 


MISCELLANEOUS  EXERCISES 


205 


37.  The  radius  of  the  earth  is  approximately  4000  mi.,  and 
the  earth  makes  one  revolution  per  day.  What  is  the  speed, 
in  miles  per  hour,  of  a  point  on  the  equator?  of  a  point  whose 
latitude  is  30°  ? 

38.  Compare  the  speed  of  a  point  on  the  equator  of  the 
earth,  in  miles  per  hour,  with  the  speed  of  an  express  train 
going  60  mi.  per  hour.  Compare  it  with  the  speed  of  a  point 
on  the  rim  of  a  flywheel  2  ft.  in  diameter  that  is  making  100 
revolutions  per  second. 

39.  If  a  hollow  pipe  has  an  inside  diameter  d  and  an  outside 
diameter  D,  the  thickness  of  its  walls  t  is  equal  to  (D  —  d)/2. 
Show  that  the  area  of  the  cross  section  of  the  metal  is 

A  D-{-d     , 


Show  that  D-\-d  =  2(2)  -  t) ;  hence  show  that  A  =  'ir{Dt-  t^). 

40.  Show  that  the  area  bounded  by  two  concentric  circles  of 
radii  r  and  R  and  two  radii  of  the  larger  one,  is  equal  to  half 
the  product  of  its  altitude  (R  —  r)  and  the  sum  of  its  two 
circular  sides.     (Compare  §  191.) 

41.  Show  how  to  find  the  area  of  a  city  lot  bounded  by  two 
circular  streets  that  have  a  common  center,  and  two  of  their 


radii,  if  the  lengths  of  the  circular  arcs  are  120  ft.  and  160  ft., 
and  the  straight  line  boundaries  are  75  ft.  long. 


APPENDIX   TO   PLANE   GEOMETRY 

MAXIMA  AND   MINIMA 

226.  Definitions.  Let  P  be  a  fixed  point  upon  the  circum- 
ference of  a  given  circle  and  let  a  series  of  chords  be  drawn 
through  this  point.  The  longest  one  of  all 
these  chords  is  the  diameter  PQ  (Ex.  2,  p.  91). 

This  fact  may  be  briefly  stated  by  saying : 

Of  all  cliords  through  P,  the  diameter  PQ 
is  the  maximum  (greatest). 

Again,  of  all  regular  polygons  that  can  be 
inscribed  in  a  given  circle,  the  one  whose 
area  is  least  is  the  inscribed  equilateral  tri- 
angle, a  fact  which  may  be  stated  by  saying :    Of  all  regular 


Fig.  154 


Fig.  155 


polygons  inscribed  in  a  circle,  the  equilateral  triangle  has  the 
minimum  (least)  area,  or  simply,  is  the  minimum. 

Likewise,  of  all  the  straight  lines  that  can  be  drawn  from  a 
fixed  point  to  a  given  line,  the  perpendicular  is  the  minimum  (§  77). 

These  and  all  other  similar  considerations  constitute  the 
subject  of  maxima  and  minima  in  Geometry.  The  maximum 
of  several  quantities  is  the  greatest  among  them  ;  the  minimum 
is  the  least  among  them.  We  shall  now  state  and  prove  a  num- 
ber of  theorems  related  to  this  subject. 

206 


§227] 


MAXIMA  AND  MINIMA 


207 


227.  Theorem    I.     Of 

aJl  triangles  that  have  the 
same  two  given  sides,  that 
in  which  these  sides  in- 
clude a  right  angle  is  the 
maximum. 

Given  the  right  A  ABC 
and  any  other  A  EBC  con- 
structed upon  the   side  BC  and  having  its  side  EB 


AB. 


To  prove 
Proof.     Draw 
Then 
But 

Therefore 
Whence  also 


A  ABC  >  A  EBC. 

ED  ±  BC. 

EB>ED. 

EB  =  AB. 

AB>ED, 
A  ABC  >  A  EBC. 


§75 

Given 
Why? 
(6),  §  190 


EXERCISES 

1.  What  is  the  maximum  line  that  can  be  drawn  within  a 
rectangle  and  terminated  by  the  sides  ?  What  is  the  minimum 
line  through  a  given  interior  point  ? 

2.  Draw  a  circle  and  take  any  point  P  within  it.  Construct 
the  maximum  line  and  also  the  minimum  line  from  Pto  the 
circumference.     Repeat,  using  a  point  P  outside  the  circle. 

3.  In  the  figure,  C  is  the  cylinder  of  a  steam  engine,  P  the 
piston,  PE  the  piston  rod,  RL  the  connecting  rod,  and  IF  the 
driving  wheel.  As  the  engine  works, 
describe  the  position  of  the  connect- 
ing rod  when  the  area  of  the  triangle 
PLO  is  a  maximum.  (0  represents 
the  center  of  the  wheel  W.)  Answer 
the  same  question  for  a  minimum  triangle  RLO. 


208  APPENDIX  [§  228 

228.  Definition.     Fij^-ures  having  equal  perimeters  are  called 
isoperimetric. 

229.  Theorem  II.     Of  all  isoperimetric  triangles  having  the 
same  base,  the  isosceles  is  the  maximum. 

Given  the  isosceles  A  ABC  and  any  other  A  DBO  constructed 
upon  EC  and  isoperimetric  to  ABC. 
To  prove  A  ABC  >  A  DBC. 


Fig.  157 

Proof.     Prolong  BA  to  E,  making  AE  =  BA,  and  draw  EC. 

A  circle  having  A  as  center  and  BE  as  diameter  can  now  be 

drawn  through  the  points  B,  C,  and  E.  Why  ? 

Therefore  Z  BCE  is  a  right  angle.  Why  ? 

Now  draw  AG  and  DF  II  BC 

also  draw  DH  =  DC,  and  join  B  and  H. 

Then      AB  + AG=DB  +  DC=  DB-\- DH=BE.  Given 

But                             DB  +  DH  >  BH.  Why  ? 

Therefore  BE  >  BH. 

Whence  also                         CE  >  CIT.  Why  ? 

But                    CG  =  \CE  and  (7i<^=  i  C^.  Why  ? 

Therefore                               CG^  >  CF-,  Why? 

and  hence                           A  ABC  >  A  Z)J5(7.  Why  ? 

230.   Corollary  1.     Of  all  isoperimetric  triangles,  the  equilateral 
is  the  maximum. 


§231]  MAXIMA  AND  MINIMA  209 

231.   Theorem  III.     Of  all  isoperimetric  polygons  having  the 
same  number  of  sides,  the  maximum  is  the  one  that  is  equilateral. 

F      A 


Given  the  maximum  polygon  ABODE  of  all  those  that  can 
be  drawn  isoperimetric  to  each  other,  having  the  same  number 
of  sides. 

To  prove  that  ABODE  is  equilateral. 

Proof.  If  ABODE  is  not  equilateral,  at  least  two  of  its 
sides,  as  AB  and  AE,  must  be  unequal ;  and  we  may  construct 
on  the  diagonal  BE  an  isosceles  A  BFE  which  is  isoperimetric 
with  A  ABE. 

Then  A  BFE  >  A  ABE.  §  229 

Therefore  BODEF  >  ABODE. 

But  this  is  contrary  to  hypothesis. 

Therefore,  AB  =  AE  and  ABODE  is  equilateral. 

EXERCISES 

1.  Show  that  of  all  triangles  inscribed  in  a  semicircle,  that 
is  greatest  which  has  the  diameter  as  base  and  the  radius,  r^ 
as  altitude.     Its  area  is  equal  to  r\ 

2.  Compare  the  area  of  the  triangle  of  Ex.  1  with  that  of  the 
semicircle ;  with  the  area  of  a  square  inscribed  in  the  circle. 

3.  Show  that  of  all  isoperimetric  quadrilaterals,  the  maxi- 
mum is  a  rhombus ;  show  furthermore  that  it  is  a  square. 

4.  Prove  that  of  all  parallelograms  having  given  sides,  the 
rectangle  is  the  maximum.  What  does  this  theorem  become 
when  stated  with  reference  to  jointed  frames  ?   (See  Ex.  7,  p.  165.) 


210  APPENDIX  [§232 

232.  Theorem  IV.  Of  all  x>olygons  with  sides  all  given  but 
one,  the  maximum  (in  area)  can  be  inscribed  in  a  semicircle 
having  the  undetermined  side  for  its  diameter. 


C 


Fig.  159 


Given  the  lengths  of  all  sides  of  the  polygon  ABCDE, 
except  EA ;  and  given  that  ABCDE  is  the  maximum  polygon 
(in  area)  that  has  the  given  sides  and  any  other  side  EA. 

To  prove  that  ABCDE  can  be  inscribed  in  a  semicircle  whose 
diameter  is  the  remaining  side,  EA. 

Proof.    From  any  vertex,  as  C,  draw  CA  and  CE. 

The  A  ACE  must  be  the  maximum  of  all  A  having  the 
given  sides  CA,  CE;  otherwise,  by  increasing  or  diminishing 
the  /.ACE,  meanwhile  keeping  the  lengths  of  CA,  CE  un- 
changed, as  also  the  form  of  the  figures,  ABC,  CDE,  but  allow- 
ing A  and  E  to  slide  along  MN  while  C  is  lowered  or  raised, 
we  can  increase  the  A  ACE,  while  the  rest  of  the  polygon 
remains  unchanged  in  area.  Hence,  unless  A  ACE  is  the 
maximum,  we  can  by  such  processes  increase  the  area  of 
ABCDE.  To  increase  in  this  way  the  area  of  ABCDE  is, 
however,  to  deny  the  hypothesis  that  ABCDE  is  the  maximum 
polygon.  Hence,  the  A  ACE  is  the  maximum  that  can  be 
drawn,  having  the  sides  AC,  CE. 

Therefore,  A  ACE  is  a  right  triangle.  §  227 

Hence,  also,  (7  lies  on  the  semicircumference  of  which  AE  is 
diameter.  Why  ? 

Likewise,  every  vertex  can  be  shown  to  lie  on  the  semicir- 
cumference whose  diameter  is  AE-,  that  is,  the  maximum 
polygon  can  be  inscribed  as  stated  in  the  theorem. 


§234] 


MAXIMA  AND   MINIMA 


211 


233.   Theorem  V.     Of  all  polygons  ivith  the  same  given  sides, 
that  ivhich  can  be  inscnbed  in  a  circle  is  the  maximum. 


B^^_      ^^^^  B'  ^C' 

Fig.  160 

Given  a  polygon  ABODE  inscribed  in  a  circle  and  mutually 
equilateral  to  another  given  polygon,  A'B'O'jyE',  which  can- 
not be  inscribed. 

To  prove  ABODE  >  A'BfCUE', 

Proof.  From  A  draw  a  diameter  AF  and  join  F  to  the 
adjacent  vertices,  D  and  O.  On  0'D\=OD)  construct 
AO'F'D'^ACDF  and  draw  AF'. 

Then,      AEDF  >  AE'DF'  and  ABOF  >  A'B'O'F'.     §  232 

Adding  these  two  inequalities, 

ABOFDE  >  A'B'O'F'D'E'. 

Take  away  from  the  two  figures  the  equal  A  OFD,  CF'  U , 
and  we  have,  ABODE  >  A'B'CD'E'. 

234.  Corollary  1 .  Of  all  isoperimetric  polygons  of  a  given 
number  of  sides,  the  maximum  is  regular. 

EXERCISE 
1.  If  in  the  figure  of  §  232  we  regard  the  sides  AB,  BO, 
etc.,  as  stiff  rods  and  assume  that  the  rod  AB  is  attached  to 
the  rod  BO  by  means  of  a  hinge  at  B,  with  a  similar  arrange- 
ment at  each  of  the  joints,  we  have  what  is  known  as  a  jointed 
frame.  Considering  all  the  different  forms  which  this  jointed 
frame  ABODE  can  assume,  what  can  be  said  of  that  one 
whose  area  is  the  maximum  ? 


212 


APPENDIX 


[§235 


235.    Theorem  VI.     Of  tivo  isoperimetric  regular  polygojis, 
the  one  having  the  greater  number  of  sides  has  the  greater  area. 


Fig.  161 

Given  a  regular  polygon  of  three  sides  (equilateral  triangle) 
ABO  and  a  regular  polygon  of  four  sides  (square)  Q,  and  let 
ABO  and  Q  be  isoperimetric. 

To  prove  that  Q  >  ABO. 

Proof.  Draw  OD  from  O  to  any  point  in  AB,  and  upon  OD 
construct  a  triangle  ^Z> (7  which  shall  be  congruent  to  ADO'^ 
that  is,  such  that  ED  =  AO  and  EO^AD. 

The  figure  DBOE  thus  formed  is  an  irregular  polygon  of 
four  sides  which  by  construction  has  the  same  perimeter  as 
ABO,  and  hence  the  same  as  Q.  Also,  it  has  by  construction 
the  same  area  as  ABO. 

But  Q  >  the  irregular  polygon  DBOE.  §  231 

Whence  Q  >  ABO. 

In  like  manner  it  can  be  shown  that  a  regular  polygon  of 
five  sides  is  greater  than  an  isoperimetric  square,  and  so  on. 

236.  Corollary  1.  The  circle  is  the  maximum  of  all  isoperi- 
metric plane  closed  figures. 


EXERCISES 

1.  Give  the  proof  that  a  regular  polygon  of  five  sides  is 
greater  than  a  corresponding  isoperimetric  square. 

2.  Show  that  a  round  can  will  hold  more  than  a  square  can 
of  the  same  perimeter,  the  two  cans  being  of  equal  height. 


§238] 


MAXIMA  AND  MINIMA 


213 


237.  Theorem  VII.  Of  all  regular  polygons  of  the  same 
area,  that  which  has  the  greatest  number  of  sides  has  the  minimum 
perimeter. 


Q 

q" 

Fig.  1G2 


Given  the  equivalent  regular  polygons  Q  and  Q',  of  which  Q' 
has  the  greater  number  of  sides. 

To  prove  that  the  perimeter  of  Q'  >  the  perimeter  of  Q. 

Proof.  Construct  a  regular  polygon  ^'  having  the  same 
perimeter  as  ^  and  the  same  number  of  sides  as  Q. 

Then  Q'  >  Q".     Therefore  Q  >  Q'\ 

Whence  the  perimeter  of  Q>  the  perimeter  of  Q". 

But  the  perimeter  of  Q'  =  the  perimeter  of  Q". 

Therefore  the  perimeter  of  Q  >  perimeter  of  (^. 


§  235 
§182 
Cons. 


238.   Corollary  1.     Of  all  plane  closed  figures  that  are  equal 
in  arettf  the  circle  has  the  minimum  perimeter. 


MISCELLANEOUS   EXERCISES.      MAXIMA  AND  MINIMA 

1.  From  two  given  points  on  the  circumference  of  a  circle  to 
draw  two  lines  meeting  on  a  tangent  to  the  circle  and  making 
a  maximum  angle  with  each  other. 

2.  To  inscribe  the  maximum  rectangle  in  a  circle. 

3.  To  inscribe  the  maximum  rectangle  in  a  quadrant. 

4.  Prove  that  of  all  triangles  having  the 
same  base  and  area,  the  isosceles  triangle 
has  the  minimum  perimeter. 

[Hint.  Prolong  AB  to  E,  making  AE  —  AC 
Draw  EB  and  AB.  Prove  that  t\ACB^t\  AEB 
so  that  BG  -  BE.    Now  use  3,  §  28.]  B' 


./' 


TABLES 


TABLE   I 

Ratios  of  the  Sides  of  Right  Triangles 

and 

Chords  and  Arcs  of  a  Unit  Circle 

TABLE   II 

Squares  and  Square  Roots  of  Numbers 
Cubes  and  Cube  Roots  of  Numbers 

TABLE   III 

Values  of  Important  Numbers 

including 

Units  of  Measurement 


TABLE   I 

RATIOS  OF   THE   SIDES  OF   RIGHT   TRIANGLES 

AND 

LENGTHS  OF  CHORDS  AND  ARCS  OF  A  UNIT  CIRCLE 

EXPLANATION   OF   TABLE   I 

1.  Ratios  of  the  Sides  of  Right  Triangles.  If  an  angle 
given  in  the  Angle  Column  is  one  acute  angle  of  a  right  triangle: 

The  Sine  Column  gives  the  ratio  of  the  side  opposite  the  angle 
to  the  hypotenuse ; 

The  Tangent  Column  gives  the  ratio  of  the  side  opposite  the 
angle  to  the  side  adjacent  to  the  angle. 

To  find  the  Cosine  of  any  angle,  take  the  sine  of  the  comple- 
ment of  that  angle. 

2.  Chords  and  Arcs  of  a  Unit  Circle.  If  an  angle  given  in 
the  Angle  Column  is  an  angle  at  the  center  of  a  circle  of  unit 
radius : 

The  Chord  Column  gives  the  length  of  the  chord  that  subtends 
that  angle ; 

The  Arc  Column  gives  the  length  of  the  arc  that  subtends  that 
angle. 

To  find  the  lengths  of  chords  or  arcs  of  any  circle  of  radius  r, 
multiply  the  values  given  in  the  table  by  that  radius. 

The  table  is  limited  to  angles  less  than  90° ;  but  to  find  the 
chord  that  subtends  an  obtuse  angle,  first  take  half  the  angle,  find 
the  sine  of  this  half  angle,  and  multiply  by  2.  This  follows 
from  the  fact  that  the  chord  of  any  angle  is  twice  the  sine  oj 
half  that  angle. 


1] 

Quantities  Determined  by  a  Given  Angle 

iii 

Angle 

Sine 

Tan- 
gent 

Chord 

Arc 

Angle 

Sine 

Tan- 
gent 

Chord 

Arc 

0°00' 

.0000 

.0000 

.0000 

.0000 

9°  00' 

.1564 

.1584 

.1569 

.1.571 

10 

.0029 

.0029 

.0029 

.0029 

10 

.1593 

.1614 

.1598 

.1600 

20 

.0058 

.0058 

.0058 

.0058 

20 

.1622 

.1()44 

.1627 

.1629 

30  ' 

.0087 

.0087 

.0087 

.0087 

:30 

.1650 

.1673 

.1656 

.1658 

40 

.0116 

.0116 

.0116 

.0116 

40 

.1679 

.1703 

.1685 

.1687 

50 

.0145 

.0145 

.0115 

.0145 

50 

.1708 

.1733 

.1714 

.1716 

1°00' 

.0175 

.0175 

.0175 

.0175 

10°  00' 

.1736 

.1763 

.1743 

.1745 

10 

.0204 

.0204 

.0204 

.0204 

10 

.1765 

.1793 

.1772 

.1774 

20 

.02:j;3 

.0233 

.0233 

.0233 

20 

.1794 

.1823 

.1801 

.1804 

30 

.0202 

.0262 

.0262 

.0262 

30 

.1822 

.1853 

.18.-30 

.1833 

40 

.0291 

.0291 

.0291 

.0291 

40 

.1851 

.1883 

.18.59 

.18(32 

50 

.0320 

.0320 

.0320 

.0320 

50 

.1880 

.1914 

.1888 

.1891 

2°  00' 

.o;m9 

.0M9 

.0349 

.0.349 

11°00' 

.1908 

.1944 

.1917 

.1920 

10 

.0378 

.0378 

.0378 

.0378 

10 

.1937 

.1974 

.1946 

.1949 

20 

.0407 

.0407 

.0407 

.0407 

20 

.1965 

.2004 

.1975 

.1978 

m 

.04;3<j 

.0437 

.0436 

.0436 

30 

.19i)4 

.20a5 

.2004 

.2007 

40 

.04<)5 

.046(; 

.0465 

.0465 

40 

.2022 

.206.5 

.2033 

.20:36 

50 

.0494 

.0495 

.0494 

.0495 

50 

.2051 

.2095 

.2062 

.2065 

3°  00' 

.0523 

.0524 

.0524 

.0524 

12°  00' 

.2079 

.2126 

.2091 

.2094 

10 

.0552 

.0553 

.0553 

.0553 

10 

.2108 

.215(i 

.2119 

.2123 

20 

.0581 

.0582 

.0582 

.0582 

20 

.2136 

.2186 

.2148 

.21.53 

'M 

.0610 

.0612 

.0611 

.0611 

30 

.2164 

.2217 

.2177 

.2182 

40 

.0640 

.0(>41 

.0640 

.0f)40 

40 

.2193 

.2247 

.220(i 

.2211 

50 

.0669 

.0670 

.0669 

.0069 

50 

.2221 

.2278 

.22(35 

.2240 

4°  00' 

.0098 

.0699 

.0698 

.0698 

13°  00' 

.2250 

.2309 

.2264 

.2269 

10 

.0727 

.0729 

.0727 

.0727 

10 

.2278 

.2339 

.2293 

.2298 

20 

.0756 

.0758 

.0756 

.0756 

20 

.2306 

.2370 

.2322 

.2327 

30 

.0785 

.0787 

.0785 

.0785 

30 

.2.3;^ 

.2401 

.2351 

.23,56 

40 

.0814 

.0816 

.0814 

.0814 

40 

.2363 

.2432 

.2380 

.2385 

50 

.0843 

.0846 

.0843 

.0844 

50 

.2391 

.2462 

.2409 

.2414 

5°  00' 

.0872 

.0875 

.0872 

.0873 

14° 00' 

.2419 

.2493 

.2437 

.2443 

10 

.OiWl 

Mm 

.0901 

.0902 

10 

.2447 

.2524 

.2466 

.2473 

20 

.m)29 

.0934 

.0931 

.0931 

20 

.2476 

.2555 

.2495 

.2502 

■30 

.0958 

.0963 

.0960 

.0<)()0 

;io 

.2504 

.2586 

.2524 

.2531 

40 

.0987 

.091  >2 

.0989 

.0989 

40 

.2532 

.2617 

.2553 

.2560 

.W 

.1016 

.1022 

.1018 

.1018 

50 

.2560 

.2648 

.2582 

.2589 

6°  00' 

.1045 

.1051 

.1047 

.1047 

15°00' 

.2588 

.2679 

.2611 

.2618 

10 

.1074 

.1080 

.1076 

.1076 

10 

.2616 

.2711 

.2639 

.2647 

20 

.1103 

.1110 

.1105 

.1105 

20 

.2644 

.2742 

.2668 

.2676 

30 

.1132 

.1139 

.1134 

.1134 

30 

.2672 

.2773 

.2697 

.2705 

40 

.1161 

.1169 

.1163 

.1164 

40 

.2700 

.2805 

.2726 

.2734 

50 

.1190 

.1198 

.1192 

.1193 

50 

.2728 

.2836 

.2755 

.27(53 

7^  00' 

.1219 

.1228 

.1221 

.1222 

16°  00' 

.2756 

.2867 

.2783 

.2793 

10 

.1248 

.1257 

.1250 

.1251 

10 

.2784 

.2899 

.2812 

.2822 

20 

.1276 

.1287 

.1279 

.1280 

20 

.2812 

.2<)31 

.2841 

.2851 

;% 

.1305 

.1317 

.1308 

.130<) 

;30 

.2840 

.2962 

.2870 

.2880 

40 

.1334 

.134^) 

.1337 

.1338 

40 

.28(Wi 

.2^)94 

.2899 

.2i)09 

50 

.13()3 

.1376 

.13()0 

.1367 

50 

.2896 

.3026 

.2927 

.29(38 

8°  00' 

.1392 

.1405 

.1395 

.139(> 

17°  00' 

.2924 

.3057 

.2956 

.2967 

10 

.1421 

.1435 

.1424 

.1425 

10 

.2952 

.3089 

.2985 

.299(3 

20 

.1449 

.14(;5 

.1453 

.1454 

20 

.2979 

.3121 

.3014 

.3025 

30 

.1478 

.1495 

.1482 

.1484 

m 

.3007 

.3153 

.3042 

..3054 

40 

.1507 

.1524 

.1511 

.1513 

40 

.3035 

.3185 

.3071 

.3083 

50 

.1536 

.1554 

.1540 

.1542 

50 

.3062 

.3217 

.3100 

.3113 

1  9°  00' 

Aofyi 

.1584 

.1569 

.1571 

18°  00' 

.3090 

.3249 

.3129 

.3142 

iv 

Quantities 

Determined  by  . 

a,  Given  Angle 

[I 

Angle 

Sine 

Tan- 
gent 

Chord 

Arc 

Angle 

Sine 

Tan- 
gent 

Chord 

Arc 

18°00' 

.3090 

.3249 

.3129 

.3142 

27° 00' 

.4540 

.5095 

.4669 

.4712 

10 

.3118 

.3281 

.3157 

.3171 

10 

.456() 

.5132 

.4697 

.4741 

20 

.3145 

.3314 

.3186 

.3200 

20 

.4592 

.5169 

.4725 

.4771 

30 

.3173 

.3346 

.3215 

.3229 

30 

.4617 

.5206 

.4754 

.4800 

40 

.3201 

.3378 

.3244 

.3258 

40 

.4643 

.5243 

.4782 

.4829 

50 

.3228 

.3411 

.3272 

.3287 

50 

.4669 

.5280 

.4810 

.4858 

19°  00' 

.3256 

.3443 

.3301 

.3316 

28°  00' 

.4695 

.5317 

.4838 

.4887 

10 

.3283 

.3476 

.3330 

.3345 

10 

.4720 

.5354 

.4867 

.4916 

20 

.3311 

.3508 

.3358 

.3374 

20 

.4746 

.5392 

.4895 

.4945 

30 

.3338 

.3541 

.3387 

.3403 

30 

.4772 

.5430 

.4923 

.4974 

40 

.3365 

.3574 

.3416 

.3432 

40 

.4797 

.5467 

.4951 

.5003 

50 

.3393 

.3607 

.3444 

.3462 

50 

.4823 

.5505 

.4979 

.5032 

20°  00' 

.3420 

.3640 

.3473 

.3491 

29°  00' 

.4848 

.5543 

.5008 

.5061 

10 

.3448 

.3673 

.3502 

.3520 

10 

.4874 

.5581 

.5036 

.5091 

20 

.3475 

.3706 

.3530 

.3549 

20 

.4899 

..5619 

.5064 

.5120 

30 

.3502 

.3739 

.3559 

.3578 

30 

.4924 

.5658 

.5092 

.5149 

40 

.3529 

.3772 

.3587 

.3607 

40 

.4950 

.5696 

.5120 

.5178 

50 

.3557 

.3805 

.3616 

.3636 

50 

.4975 

.5735 

.5148 

.5207 

21°  00' 

.3584 

.3839 

.3645 

.3665 

30°  00' 

..5000 

.5774 

.5176 

.5236 

10 

.3611 

.3872 

.3673 

.3694 

10 

.5025 

.5812 

.5204 

.5265 

20 

.3638 

.3906 

.3702 

.3723 

20 

.5050 

.5851 

.5233 

.5294 

30 

.3665 

.3939 

.3730 

.3752 

30 

.5075 

.5890 

.5261 

.5323 

40 

.3692 

.3973 

.3759 

.3782 

40 

.5100 

.5930 

.5289 

.5352 

50 

.3719 

.4006 

.3788 

.3811 

50 

.5125 

.5969 

.5317 

.5381 

22°  00' 

.3746 

.4040 

.3816 

.3840 

31°00' 

.5150 

.6009 

.5345 

.5411 

10 

.3773 

.4074 

.3845 

.3869 

10 

.5175 

.6048 

.5373 

.5440 

20 

.3800 

.4108 

.3873 

.3898 

20 

.5200 

.6088 

.5401 

.5469 

30 

.3827 

.4142 

.3902 

.3927 

30 

.5225 

.6128 

.5429 

.5498 

40 

.3854 

.4176 

.3930 

.3956 

40 

.5250 

.6168 

.5457 

.5527 

50 

.3881 

.4210 

.3959 

.3985 

50 

.5275 

.6208 

.5485 

.5556 

23° 00' 

.3907 

.4245 

.3987 

.4014 

32°  00' 

.5299 

.6249 

.5513 

.5585 

10 

.3934 

.4279 

.4016 

.4043 

10 

.5324 

.6289 

.5541 

.5614 

20 

.3961 

.4314 

.4044 

.4072 

20 

.5348 

.6330 

.5569 

..5643 

30 

.3987 

.4348 

.4073 

.4102 

30 

.5373 

.6371 

.5597 

.5672 

40 

.4014 

.4383 

.4101 

.4131 

40 

.5398 

.6412 

.5625 

.5701 

50 

.4041 

.4417 

.4130 

.4160 

50 

.5422 

.6453 

.5652 

.5730 

24° 00' 

.4067 

.4452 

.4158 

.4189 

33°  00' 

.5446 

.6494 

.5680 

.5760 

10 

.4094 

.4487 

.4187 

.4218 

10 

.5471 

.6536 

.5708 

.5789 

20 

.4120 

.4522 

.4215 

.4247 

20 

.5495 

.6577 

.5736 

.5818 

30 

.4-147 

.4557 

.4244 

.4276 

30 

.5519 

.6619 

.5764 

.5847 

40 

.4173 

.4592 

.4272 

.4305 

40 

.5544 

.6661 

.5792 

.5876 

50 

.4200 

.4628 

.4300 

.4334 

50 

.5568 

.6703 

.5820 

.5905 

25°  00' 

.4226 

.4663 

.4329 

.4363 

34° 00' 

.5592 

.6745 

.5847 

.5934 

10 

.4253 

.4699 

.4357 

.4392 

10 

.5616 

.6787 

.5875 

.5963 

20 

.4279 

.4734 

.4386 

.4422 

20 

.5640 

.6830 

.5903 

.5992 

30 

.4305 

.4770 

.4414 

.4451 

30 

.5664 

.6873 

.5931 

.6021 

40 

.4331 

.4806 

.4442 

.4480 

40 

.5688 

.6916 

.5959 

.6050 

50 

.4358 

.4841 

.4471 

.4509 

50 

.5712 

.6959 

.5986 

.6080 

26°  00' 

.4384 

.4877 

.4499 

.4538 

35°  00' 

.5736 

.7002 

.6014 

.6109 

10 

.4410 

.4913 

.4527 

.4567 

10 

.5760 

.7046 

.6042 

.6138 

20 

.4436 

.4950 

.4556 

.4596 

20 

.5783 

.7089 

.6070 

.6167 

30 

.4462 

.4986 

.4584 

.4625 

30 

.5807 

.7133 

.6097 

.6196 

40 

.4488 

.5022 

.4612 

.4654 

40 

.5831 

.7177 

.6125 

.6225 

50 

.4514 

.5059 

.4641 

.4683 

50 

.5854 

.7221 

.6153 

.6254 

27°  00' 

.4540 

.5095 

.4669 

.4712 

36°  00' 

.5878 

.7265 

.6180 

.6283 

1] 

Quantities 

Determined  by  a  Giyen  Angle 

V 

Angle 

Sine 

Tan- 
gent 

Chord 

Arc 

Angle 

Sine 

Tan- 
gent 

Chord 

A.C 

36°  00' 

.5878 

.7265 

.6180 

.6283 

45° 00' 

.7071 

1.0000 

.7654 

.7854 

10 

.5901 

.7310 

.6208 

.6312 

10 

.7092 

1.0058 

.7681 

.7883 

20 

.5i)25 

.7355 

.6236 

.6341 

20 

.7112 

1.0117 

.7707 

.7912 

30 

.5948 

.7400 

.6263 

.6370 

30 

.7133 

1.0176 

.7734 

.7941 

40 

.5972 

.7445 

.62^)1 

.6400 

40 

.7153 

1.0235 

.7761 

.7970 

50 

.5995 

.7490 

.6318 

.6429 

50 

.7173 

1.0295 

.7788 

.7999 

37°  00' 

.6018 

.7536 

.6M6 

.6458 

46°  00' 

.7193 

1.0355 

.7815 

.8029 

10 

.OOil 

.7581 

.6374 

.(J487 

10 

.7214 

1.0416 

.7841 

.8058 

20 

.6065 

.7627 

.6401 

.6516 

20 

.72^4 

1.0477 

.7868 

.8087 

30 

.6088 

.7673 

.(>42i) 

.6545 

30 

.7254 

1.0538 

.7895 

.8116 

40 

.6111 

.7720 

.6456 

.6574 

40 

.7274 

1.0599 

.7922 

.8145 

50 

.6134 

.7766 

.6484 

.6603 

50 

.7294 

1.0661 

.7&48 

.8174 

38°  00' 

.6157 

.7813 

.6511 

.6632 

47°  00' 

.7314 

1.0724 

.7975 

.8203 

10 

.6180 

.78(50 

.a539 

.6661 

10 

.7333 

1.0786 

.8002 

.8232 

20 

.6202 

.7907 

.(5566 

.6690 

20 

.7353 

1.0850 

.8028 

.8261 

30 

.6225 

.7954 

.6594 

.6720 

30 

.7373 

1.0913 

.8055 

.82<X) 

40 

.6248 

.8002 

.6621 

.6749 

40 

.7392 

1.0977 

.8082 

.8319 

50 

.6271 

.8050 

.em9 

.6778 

50 

.7412 

1.1041 

.8108 

.8348 

39°  00' 

.6293 

.8098 

.6676 

.6807 

48°  00' 

.7431 

1.1106 

.8135 

.8378 

10 

.6316 

.8144) 

.6704 

.6836 

10 

.7451 

1.1171 

.8161 

.8407 

20 

.6338 

.8195 

.6731 

.6865 

20 

.7470 

1.1237 

.8188 

.843(5 

30 

.6361 

.8243 

.6758 

.68i>4 

30 

.7490 

1.1303 

.8214 

.8465 

40 

.6383 

.8292 

Sim 

.(5923 

40 

.750*) 

1.1369 

.8241 

.8494 

50 

.6406 

.im2 

.6813 

.6952 

50 

.7528 

1.1436 

.8267 

.8523 

40°  00' 

.6428 

.8391 

.6840 

.(5981 

49°  00' 

.7547 

1.1504 

.8294 

.8552 

10 

.(M50 

.8441 

.6868 

.7010 

10 

.75(36 

1.1571 

.8320 

.8581 

20 

.6472 

.8491 

.6895 

.7039 

20 

.7585 

1.1640 

.8347 

.8(510 

30 

.6494 

.8541 

.6922 

.7069 

30 

.7604 

1.1708 

.8373 

.8(539 

40 

.6517 

.8591 

.69.50 

.7098 

40 

.7623 

1.1778 

.8400 

.8668 

50 

.6539 

.8642 

.69/7 

.7127 

50 

.7642 

1.1847 

.8426 

.8698 

41°  00' 

.6561 

.8693 

.7004 

.7156 

60°  00' 

.7(560 

1.1918 

.8452 

.8727 

10 

.65a3 

.8744 

.7031 

.7185 

10 

.7679 

1.1988 

.8479 

.8756 

20 

.(5604 

.8796 

.7059 

.7214 

20 

.7698 

1.2059 

.8505 

.8785 

30 

.6626 

.8847 

.7086 

.7243 

30 

.7716 

1.2131 

.8531 

.8814 

40 

.(i648 

.8899 

.7113 

.7272 

40 

.7735 

1.2203 

.8558 

.8843 

50 

.f;670 

.8952 

.7140 

.7301 

50 

.7753 

1.2276 

.8584 

.8872 

42°  00' 

.6691 

.9004 

.7167 

.7330 

51°  00' 

.7771 

1.2349 

.8610 

.8901 

10 

.6713 

.9057 

.7195 

.7359 

10 

.7790 

1.2423 

.8636 

.8930 

20 

.6734 

.9110 

.7222 

.7389 

20 

.7808 

1.2497 

.8663 

.8959 

30 

.6756 

.9163 

.7249 

.7418 

30 

.7826 

1.2572 

.8689 

.8988 

40 

.6VVV 

.9217 

.7276 

.7447 

40 

.7844 

1.2647 

.8715 

.9018 

50 

.6799 

.9271 

.7303 

.7476 

50 

.7862 

1.2723 

.8741 

.9047 

43°  00' 

M20 

.9325 

.7330 

.7505 

62°  00' 

.7880 

1.2799 

.8767 

.9076 

10 

.6841 

.9380 

.7357 

.7634 

10 

.7898 

1.2876 

.8794 

.9105 

20 

.6862 

.9435 

.7384 

.7563 

20 

.7916 

1.2954 

.8820 

.9134 

30 

MM 

.9490 

.7411 

.7592 

m 

.79:34 

1.3032 

.884(> 

.9163 

40 

.6905 

.9545 

.7438 

.7621 

40 

.7951 

1.3111 

.8872 

.9192 

50 

.6926 

.m)i 

.7465 

.76.'>0 

50 

.79(59 

1.3190 

.8898 

.9221 

44°  00' 

.6947 

.9657 

.7492 

.7(579 

53°  00' 

.7986 

1.3270 

.8924 

.9250 

10 

.6967 

.9713 

.7519 

.7709 

10 

.8004 

1.3351 

.8950 

.9279 

20 

.6988 

.9770 

.7546 

.7738 

20 

.8021 

1.3432 

.8976 

.9308 

30 

.7009 

.9827 

.7573 

.7767 

30 

.80.39 

1.3514 

.9002 

.9338 

40 

.7030 

.9884 

.7600 

.7796 

40 

.8056 

1.3597 

.9028 

.9367 

50 

.7050 

.^1942 

.7627 

.7825 

50 

.8073 

1.3680 

.9054 

.9396 

45°  00' 

.7071 

1.0000 

.7654 

.7854 

54°  00' 

.80i)0 

1.3764 

.9080 

.9425 

vi 

Quantities  Determined  by  . 

a  Griven  Angle 

[1 

Angle 

Sine 

Tan- 
gent 

Chord 

Arc 

Angle 

Sine 

Tan- 
gent 

Chord 

Arc 

54°  00' 

.8090 

1.3764 

.9080 

.9425 

63°  00' 

.8910 

1.9626 

1.0450 

1.0996 

10 

.8107 

1.3848 

.9106 

.9454 

10 

.8923 

1.9768 

1.0475 

1.1025 

20 

.8124 

1.39M 

.9132 

.9483 

20 

.8936 

1.9912 

1.0500 

1.1054 

30 

.8141 

1.4019 

.9157 

.9512 

30 

.8949 

2.0057 

1.0524 

1.1083 

40 

.8158 

1.4106 

.9183 

.9541 

40 

.8962 

2.0204 

1.0549 

1.1112 

50 

.8175 

1.4193 

.9209 

.9570 

50 

.8975 

2.0353 

1.0574 

1.1141 

65°00' 

.8192 

1.4281 

.9235 

.9599 

64°  00' 

.8988 

2.0503 

1.0598 

1.1170 

10 

.8208 

1.4370 

.9261 

.9628 

10 

.9001 

2.0655 

1.0623 

1.1199 

20 

.8225 

1.4460 

.9287 

.9657 

20 

.i)013 

2.0809 

1.0648 

1.1228 

30 

.8241 

1.4550 

.9312 

.9687 

30 

.9026 

2.0965 

1.0672 

1.1257 

40 

.8258 

1.4641 

.9338 

.9716 

40 

.9038 

2.1123 

1.0697 

1.1286 

50 

.8274 

1.4733 

.9364 

.9745 

50 

.9051 

2.1283 

1.0721 

1.1316 

56°  00' 

.8290 

1.4826 

.9389 

.9774 

65°  00' 

.9063 

2.1445 

1.0746 

1.1345 

10 

.8307 

1.4919 

.9415 

.9803 

10 

.9075 

2.1609 

1.0771 

1.1374 

20 

.8323 

1.5013 

.9441 

.9832 

20 

.9088 

2.1775 

1.0795 

1.1403 

30 

.8339 

1.5108 

.9466 

.9861 

30 

.9100 

2.1943 

1.0820 

1.1432 

40 

.8355 

1.5204 

.9492 

.9890 

40 

.9112 

2.2113 

1.0844 

1.1461 

50 

.8371 

1.5301 

.9518 

.9919 

50 

.9124 

2.2286 

1.0868 

1.1490 

57° 00' 

.8387 

1.5399 

.9543 

.9948 

66°  00' 

.9135 

2.2460 

1.0893 

1.1519 

10 

.8403 

1.5497 

.9569 

.9977 

10 

.9147 

2.2637 

1.0917 

1.1548 

20 

.8418 

1.5597 

.9594 

1.0007 

20 

.9159 

2.2817 

1.0942 

1.1577 

30 

.8434 

1.5697 

.9620 

1.0036 

30 

.9171 

2.2998 

1.09()6 

1.1606 

40 

.8450 

1.5798 

.9645 

1.0065 

40 

.9182 

2.3183 

1.0990 

1.1636 

50 

.8465 

1.5900 

.9671 

1.0094 

50 

.9194 

2.3369 

1.1014 

1.1665 

58°  00' 

.8480 

1.6003 

.9696 

1.0123 

67°  00' 

.9205 

2.3559 

1.1039 

1.1691 

10 

.84% 

1.6107 

.9722 

1.0152 

10 

.9216 

2.3750 

1.1063 

1.1723 

20 

.8511 

1.6212 

.9747 

1.0181 

20 

.9228 

2.3945 

1.1087 

1.1752 

30 

.8526 

1.6319 

.9772 

1.0210 

30 

.9239 

2.4142 

1.1111 

1.1781 

40 

.8542 

1.6426 

.i)798 

1.0239 

40 

.9250 

2.4342 

1.1136 

1.1810 

50 

.8557 

1.6534 

.9823 

1.0268 

50 

.9261 

2.4545 

1.1160 

1.1839 

59°  00' 

.8572 

1.6643 

.9848 

1.0297 

68°  00' 

.9272 

2.4751 

1.1184 

1.1808 

10 

.8587 

1.6753 

.9874 

1.0327 

10 

.9283 

2.4960 

1.1208 

1.1897 

20 

.8601 

1.6864 

.9899 

1.0356 

20 

.9293 

2.5172 

1.1232 

1.1926 

30 

.8616 

1.6977 

.9924 

1.0385 

30 

.9304 

2.5386 

1.1256 

1.1956 

40 

.8631 

1.7090 

.9950 

1.0414 

40 

.9315 

2.5605 

1.1280 

1.1985 

50 

.8646 

1.7205 

.9975 

1.0443 

50 

.9325 

2.5826 

1.1304 

1.2014 

60°  00' 

.8660 

1.7321 

1.0000 

1.0472 

69°  00' 

.9336 

2.6051 

1.1328 

1.2043 

10 

.8675 

1.7437 

1.0025 

1.0501 

10 

.9346 

2.6279 

1.1352 

1.2072 

20 

.8689 

1.7556 

1.0050 

1.0530 

20 

.9356 

2.6511 

1.1376 

1.2101 

30 

.8704 

1.7675 

1.0075 

1.0559 

30 

.9367 

2.6746 

1.1400 

1.2130 

40 

.8718 

1.7796 

1.0101 

1.0588 

40 

.9377 

2.6985 

1.1424 

1.2159 

50 

.8732 

1.7917 

1.0126 

1.0617 

50 

.9387 

2.7228 

1.1448 

1.2188 

61°  00' 

.8746 

1.8040 

1.0151 

1.0647 

70°  00' 

.9397 

2.7475 

1.1472 

1.2217 

10 

.8760 

1.8165 

1.0176 

1.0676 

10 

.9407 

2.7725 

1.1495 

1.2246 

20 

.8774 

1.8291 

1.0201 

1.0705 

20 

.9417 

2.7980 

1.1519 

1.2275 

30 

.8788 

1.8418 

1.0226 

1.0734 

30 

.9426 

2.8239 

1.1543 

1.2305 

40 

.8802 

1.8546 

1.0251 

1.0763 

40 

.9436 

2.8502 

1.1567 

1.2334 

50 

.8816 

1.8676 

1.0276 

1.0792 

.50 

.9446 

2.8770 

1.1590 

1.2363 

62°  00' 

.8829 

1.8807 

1.0301 

1.0821 

71°00' 

.9455 

2.9042 

1.1614 

1.2392 

10 

.8843 

1.8940 

1.0326 

1.0850 

10 

.9465 

2.9319 

1.1638 

1.2421 

20 

.8857 

1.9074 

1.0351 

1.0879 

20 

.9474 

2.9(500 

1.1661 

1.2450 

30 

.8870 

1.9210 

1.0375 

1.0908 

30 

.9483 

2.9887 

1.1685 

1.2479 

40 

.8884 

1.9347 

1.0400 

1.0937 

40 

.9492 

3.0178 

1.1709 

l.?508 

50 

.8897 

1.9486 

1.0425 

1.0966 

50 

.9502 

3.0475 

1.1732 

1.2537 

63° 00' 

.8910 

1.9626 

1.0450 

1.0996 

72°  00' 

.9511 

3.0777 

1.1756 

1.2566 

1] 

Quantities  Determined  by  a  Given  Anj^le 

vii 

Angle 

Sine 

Tan- 
gent 

Chord 

Arc 

Angle 

Sine 

Tan- 
gent 

Chord 

Arc 

72^00' 

.9511 

3.0777 

1.1756 

1.2566 

81° 00' 

.9877 

6.3138 

1.2989 

1.4137 

10 

.9520 

3.1084 

1.1779 

1.2595 

10 

.9881 

6.4348 

1.3011 

1.4166 

20 

.9528 

3.1397 

1.1803 

1.262,-. 

20 

.988(5 

6.560(5 

1.3033 

1.4195 

riO 

.9537 

3.1716 

1.1826 

1.2654 

30 

.98iK) 

6.(5912 

1.3055 

1.4224 

40 

.9546 

3.2041 

1.1850 

1.2683 

40 

.9894 

6.82(59 

1.3077 

1.4254 

oO 

.9555 

3.2371 

1.1873 

1.2712 

5(3 

.9899 

6.9682 

1.3m)9 

1.4283 

73°  00' 

.9503 

3.2709 

1.1896 

1.2741 

82°  00' 

.9903 

7.11.54 

1.3121 

1.4312 

10 

.t)572 

3.3052 

1.1920 

1.2770 

10 

.9<J()7 

7.2687 

1.3143 

1.4:341 

20 

.9580 

3.;M02 

1.1943 

1.2799 

20 

.9911 

7.4287 

1.3165 

1.4,370 

30 

.9588 

3.3759 

1.1966 

1.2828 

30 

.9^n4 

7.5958 

1.3187 

ia:m) 

40 

.95*  Hi 

3.4124 

1.1990 

1.2857 

40 

.9918 

7.7704 

1.3-209 

1.4428 

50 

.IKiOo 

3.4495 

1.2013 

1.2886 

50 

.9922 

7.9530 

1.3231 

1.4457 

74^00' 

.%13 

3.4874 

1.2036 

1.2915 

83°  00' 

.9925 

8.1443 

1.32,52 

1.4486 

10 

.%21 

3..5261 

1.2060 

1.2<>45 

10 

.9929 

8.3450 

1.3274 

1.4515 

20 

.9()28 

3. .5656 

1.2083 

1.2974 

20 

.9932 

8.5555 

1.32ii6 

1.4.544 

30 

.di'm 

3.()059 

1.2106 

1.3003 

30 

.9936 

8.7769 

1.3:318 

1.4573 

40 

.%44 

3.6470 

1.2129 

1.3032 

40 

.9939 

9.0098 

1.3339 

1.4603 

50 

.mo2 

3.6891 

1.2152 

1.3061 

50 

.9942 

9.2553 

1.33(51 

1.4632 

75° 00' 

.96,59 

3.7321 

1.2175 

1.3090 

84°  00' 

.9945 

9.5144 

l.a383 

1.4661 

10 

.9667 

3.7760 

1.2198 

1.3119 

10 

.9948 

9.7882 

1.3404 

1.4(590 

20 

.9(J74 

3.8208 

1.2221 

1.3148 

20 

.9951 

10.078 

l.:3426 

1.4719 

:^o 

.9()81 

3.86<)7 

1.2244 

1.3177 

30 

.9954 

10.385 

1.3447 

1.4748 

40 

.9689 

3.9136 

1.2267 

1.3206 

40 

.9^357 

10.712 

1.34(59 

1.4777 

r)0 

.96i)(j 

3.9617 

1.2290 

1.3235 

50 

.9959 

11.059 

IMM 

1.4806 

76^00' 

.9703 

4.0108 

1.2313 

1.3265 

85°  00' 

.9962 

11.430 

1.3512 

1.48,35 

10 

.9710 

4.0611 

1.2336 

1.3294 

10 

.91>64 

11.826 

1.3533 

1.48(54 

20 

.9717 

4.1126 

1.2359 

1.3323 

20 

.9967 

12.251 

1.3555 

1.4893 

:iO 

.9724 

4.1653 

1.2382 

1.3352 

30 

.99(59 

12.706 

1.3576 

1.4923 

40 

.97;^X) 

4.2193 

1.2405 

1.3:381 

40 

.9971 

13.197 

1.3597 

1.49,52 

50 

.9737 

4.2747 

1.2428 

1.3410 

50 

.9974 

13.727 

1.3619 

1 .4981 

77° 00' 

.9744 

4.3315 

1.2450 

1.3439 

86°  00' 

.9976 

14.301 

1.3640 

l.,5010 

10 

.9750 

4.3897 

1.2473 

1.3468 

10 

Mm 

14.924 

1.3661 

1.5039 

20 

.9757 

4.4494 

1.2496 

1.3497 

20 

.9980 

15.605 

1..3()82 

l.,5068 

30 

.9763 

4.5107 

1.2518 

1.3526 

30 

.9<)81 

16.350 

1.3704 

l.,5097 

40 

.9769 

4.5736 

1.2541 

1.3555 

40 

.9983 

17.169 

1.3725 

1.5126 

50 

.9775 

4.6382 

1.2564 

1.3584 

50 

.9985 

18.075 

1.3746 

1.5155 

78^00' 

.9781 

4.7046 

1.2586 

1.3614 

87°  00' 

.9986 

19.081 

1.3767 

1.5184 

10 

.9787 

4.7729 

1.2609 

l.:3(>43 

10 

.9988 

20.206 

1.3788 

1.5213 

20 

.9793 

4.8430 

1.2632 

1.3672 

20 

.9989 

21.470 

1..3809 

1.5243 

30 

.9799 

4.9152 

1.2654 

1.3701 

30 

.9990 

22.904 

1.3830 

1.5272 

40 

.9805 

4.9894 

1.2677 

1..3730 

40 

.9992 

24.542 

1.3851 

1.5301 

50 

.9811 

5.0658 

1.2699 

1.3759 

W 

.99f)3 

26.432 

1.3872 

1.5330 

79°  00' 

.9816 

5.144<; 

1.2722 

1.3788 

88° 00' 

.9994 

28.6:36 

1..3893 

1.5:3.59 

10 

.9822 

5.22r»7 

1.2744 

l.:i817 

10 

.<M)o 

31.242 

1.3914 

1.5:388 

20 

.9827 

5..30<»3 

1.2766 

1.3846 

20 

MM) 

a4.368 

l.:39;35 

1..5417 

30 

:9833 

5.3955 

1.2789 

1.3875 

;i0 

.mn 

:38.188 

1., -395(5 

1.5446 

40 

.9a38 

5.4^45 

1.2811 

1.3904 

40 

Mm 

42.9(54 

l.;J977 

1.5475 

50 

.9843 

5.57(;4 

1.2833 

1.3934 

50 

MM8 

49.104 

1.3997 

1.5504 

80°  00' 

.9848 

5.6713 

1.2856 

1.39(53 

89°  00' 

.9998 

57.290 

1.4018 

1.5,533 

10 

.9853 

5.7694 

1.2878 

1.3i)92 

10 

.<)99i) 

68.750 

1.40:39 

1.55(53 

20 

.9858 

5.8708 

1.2900 

1.4021 

20 

.i)999 

85.940 

1.4060 

1.. 5.592 

30 

.9863 

5.9758  1 1.2922 

1.4050 

30 

1.0000 

114.59 

1.4080 

1.. 5(521 

40 

.9868 

6.0844  1  1.2945 

1.4079 

40 

1.0000 

171.89 

1.4101 

1.5(550 

50 

.9872 

6.1970 

1.2967 

1.4108 

50 

1.0000 

343.77 

1.4122 

1.5679 

81° 00' 

.9877 

6.3138 

1.2989 

1.4137 

90°  00' 

1.0000 

1.4142 

1.5708 

TABLE  II  — POWERS  AND  ROOTS 

EXPLANATION   OF  TABLE   II 

1.  Squares  and  Cubes.  The  squares  of  numbers  between 
1.00  and  10.00  at  intervals  of  .01  are  given  in  column  headed 
n^.  To  find  the  square  of  any  other  number,  divide  (or  mul- 
tiply) the  given  number  by  10  to  reduce  it  to  a  number  between 
1  and  10;  find  the  square  of  this  last  number;  multiply  (or 
divide)  the  square  thus  found  by  10  twice  as  many  times  as 
you  did  the  given  number. 

The  cube  is  given  in  the  column  headed  n\  To  find  the  cube 
of  any  number  not  between  1  and  10,  first  reduce  that  number 
to  a  number  between  1  and  10  by  dividing  (or  multiplying)  by 
a  power  of  10.  Multiply  (or  divide)  the  result  found  by  three 
times  as  high  a  power  of  10  as  was  used  to  reduce  the  given 
number. 

2.  Square  Roots.  The  square  roots  of  numbers  between  1 
and  10  are  found  in  the  column  headed  Vn. 

The  square  roots  of  numbers  between  10  and  100  may  be 
found  in  the  column  headed  VlO  n. 

The  square  roots  of  numbers  between  100  and  1000  may  be 
found  in  the  column  headed  Vn  by  multiplying  the  given  root 
by  10,  since  VlOOn  =  10  Vn. 

Other  square  roots  may  be  found  in  a  similar  manner. 

3.  Cube  Roots,     The  column  headed : 

Vn  gives  cube  roots  of  numbers  between  1  and  10 ; 
-\/10n  gives  cube  roots  of  numbers  between  10  and  100 ; 
VlOOn  gives  cube  roots  of  numbers  between  100  and  1000. 

To  find  the  cube  root  of  a  number  between  1000  and  10000, 
take  10  times  the  value  found  in  the  column  headed  Vn,  since 
ViOOO^  =  10  -s/n. 

Other  cube  roots  may  be  found  similarly. 


viii 


11] 

Table  II  —  Powers  and  Boots 

IX 

n 

n2 

v^ 

VlOn 

n=^ 

Vn 

^10  n 

S/lOOw 

1.00 

1.0000 

1.00000 

3.16228 

1.00000 

1.00000 

2.15443 

4.64159 

1.01 
1.02 
1.03 

1.04 
1.05 
1.06 

1.07 
1.08 
1.09 

1.0201 
1.0404 
1.0609 

1.081& 
1.1025 
1.1236 

1.1449 
lA(m 
1.1881 

1.00499 
1.00995 
1.01489 

1.01980 
1.02470 
1.02956 

1.03441 
1.03923 
1.04403 

3.17805 
3.19374 
3.20936 

3.22490 
3.24037 
3.25576 

3.27109 
3.28634 
3.30151 

1.03030 
1.06121 
1.09273 

1.12486 
1.15762 
1.19102 

1.22504 
1.25971 
1.29503 

1.00332 
1.00662 
1.00990 

1.01316 
1.01640 
1.01961 

1.02281 
1.025m) 
1.02914 

2.16159 
2.16870 
2.17577 

2.18279 
2.18976 
2.19669 

2.20358 
2.21042 
2.21722 

4.65701 
4.67233 

4.687.55 

4.70267 
4.717(59 
4.73262 

4.74746 

4.76220 
4.77686 

1.10 

1.2100 

1.04881 

3.31662 

1.33100 

1.03228 

2.22398 

4.79142 

1.11 
1.12 
1.13 

1.14 
1.15 

1.16 

1.17 
1.18 
1.19 

1.2321 
1.2544 
1.2769 

1.2996 
1.3225 
1.3456 

1.3689 
1.3924 
1.4161 

1.05357 
1.05830 
1.06301 

1.06771 
1.07238 
1.07703 

1.08167 
1.08628 
1.09087 

3.33167 
3.34664 
3.36155 

3.37639 
3.39116 
3.40588 

3.42053 
3.43511 
3.44904 

1.36763 
1.40493 
1.44290 

1.48154 
1.52088 
1.5(>090 

1.C0161 
1.64303 
1.68516 

1.0a540 
1.03850 
1.04158 

1.044()4 
1.04769 
1.05072 

1.05373 
1.05672 
1.05970 

2.23070 
2.23738 
2.24402 

2.25062 
2.25718 
2.26370 

2.27019 
2.27664 
2.28305 

4.80590 
4.82028 
4.83459 

4.84881 
4.86294 
4.87700 

4.89097 
4.90487 
4.91868 

1.20 

1.4400 

1.09545 

3.40410 

1.72800 

1.062G6 

2.28943 

4.93242 

1.21 
1.22 
1.23 

1.24 
1.25 
1.26 

1.27 
1.28 
1.29 

1.4641 

1.4884 
1.5129 

1.5376 
1.5625 
1.5876 

1.6129 
1.6384 
1.6641 

1.10000 
1.10454 
1.10905 

1.11355 
1.11803 
1.12250 

1.12694 
1.13137 
1.1.3578 

3.47851 
3.49285 
3.50714 

3.52136 
3.53553 
3.54965 

3.56371 
3.57771 
3.5916(} 

1.77156 
1.81585 
1.86087 

1.90(5(32 
1.95312 
2.00038 

2.04838 
2.09715 
2.14669 

1.06560 
1.06853 
1.07144 

1.07434 
1.07722 
1.08008 

1.08293 
1.08577 
1.08859 

2.29577 
2.30208 
2.30835 

2.31459 
2.32079 
2.32697 

2.33311 
2.33921 
2.34529 

4.94609 
4.95968 
4.97319 

4.986(33 
5.00000 
5.01330 

5.02653 
5.03968 
5.05277 

1.30 

1.6900 

1.14018 

3.60,555 

2.19700 

1.09139 

2.35133 

5.06580 

1.31 
1.32 
1.33 

1.34 
1.35 
1.36 

1.37 
1.38 
1.39 

1.7161 
1.7424 
1.7689 

1.7956 
1.8225 
1.8496 

1.8769 
l.i>044 
1.9321 

1.14455 
1.14891 
1.15326 

1.15758 
1.16190 
1.16619 

1.17047 
1.17473 
1.17898 

3.61939 
3.63318 
3.64692 

3.66060 
3.67423 
3.68782 

3.70135 
3.71484 
3.7282V 

2.24809 
2.299<)7 
2.35264 

2.40610 
2.46038 
2.51546 

2.57135 
2.62807 
2.68562 

1.09418 
1.0969(5 
1.09972 

1.10247 
1.10521 
1.10793 

1.11064 
1.113:M 
1.11602 

2.35735 
2.3<i333 
2.36928 

2.37521 
2.38110 
2.38697 

2.39280 
2.398(31 
2.40439 

5.07875 
5.09164 
5.10447 

5.11723 
5.12993 
5.14256 

5.15514 
5.16765 
5.18010 

1.40 

1.9600 

1.18322 

3.74166 

2.74400 

1.11869 

2.41014 

5.19249 

1.41 
1.42 
1.43 

1.44 
1.45 
1.46 

1.47 
1.48 
1.4^ 

1.9881 
2.0164 
2.0449 

2.0736 
2.1025 
2.1316 

2.1609 
2.1904 
2.2201 

1.18743 
1.19164 
1.19583 

1.20000 
1.20416 
1.20830 

1.21244 
1.21655 
1.22066 

3.75500 
3.76829 
3.78153 

3.79473 
3.80789 
3.8209i) 

3.83406 
3.»4708 
3.8(J005 

2.80322 
2.86:^29 
2.92421 

2.9a598 
3.04862 
3.11214 

3.17652 
3.24179 
3.30795 

1.12135 
1.12399 
1.12662 

1.12924 
1.13185 
1.13445 

1.13703 
1.13960 
1.14216 

2.41587 
2.42156 
2.42724 

2.43288 
2.43850 
2.44409 

2.44966 
2.45520 
2.46072 

5.20483 
5.21710 
5.22932 

5.24148 
5.25359 
5.26564 

5.27763 
5.28957 
5.30146 

X 

Powers  and  Roots 

[11 

n 

n^ 

y/n 

VlOw, 

n^ 

^n 

^10^2, 

^100*1 

1.50 

2.2500 

1.22474 

3.87298 

3.37500 

1.14471 

2.46621 

5.31329 

1.51 
1.52 
1.53 

1.54 

1.55 
1.56 

1.57 
1.58 
1.59 

2.2801 
2.3104 
2.3409 

2.3716 
2.4025 
2.4336 

2.4649 
2.4964 

2.5281 

1.22882 
1.23288 
1.23693 

1.24097 
1.24499 
1.24900 

1.25300 
1.25698 
1.26095 

3.88587 
3.89872 
3.91152 

3.92428 
3.93700 
3.94968 

3.96232 
3.97492 
3.98748 

3.44295 
3.51181 
3.58158 

3.65226 

3.72388 
3.79642 

3.86989 
3.94431 
4.01908 

1.14725 

1.14978 
1.15230 

1.15480 
1.15729 
1.15978 

1.16225 
1.16471 
1.16717 

2.47168 
2.47712 
2.48255 

2.48794 
2.49332 
2.49867 

2.50399 
2.50930 
2.51458 

5.32507 
5.33680 
5.34848 

5.36011 
5.37169 
5.38321 

5.39469 
5.40612 
5.41750 

1.60 

2.5600 

1.26491 

4.00000 

4.09600 

1.169(31 

2.51984 

5.42884 

1.61 
1.62 
1.63 

1.64 
1.65 
1.66 

1.67 
1.68 
1.69 

2.5921 
2.6244 
2.6569 

2.6896 
2.7225 
2.7556 

2.7889 
2.8224 
2.8561 

1.20886 
1.27279 
1.27671 

1.28062 
1.28452 
1.28841 

1.29228 
1.29615 
1.30000 

4.01248 
4.02492 
4.03733 

4.049(39 
4.06202 
4.07431 

4.08056 
4.09878 
4.11096 

4.17328 
4.25153 
4.33075 

4.41094 
4.49212 
4.57430 

4.65746 
4.74163 
4.82681 

1.17204 
1.17446 

1.17687 

1.17927 

1.18167 
1.18405 

1.18642 
1.18878 
1.19114 

2.52508 
2.53030 
2.53549 

2.54067 
2.54582 
2.55095 

2.55607 
2.56116 
2.56623 

5.44012 
5.45136 
5.46256 

5.47370 
5.48481 
5.49586 

5.50688 
5.51785 

6.52877 

1.70 

2.8900 

1.30384 

4.12311 

4.91300 

1.19348 

2.57128 

5.53966 

1.71 
1.72 
1.73 

1.74 
1.75 
1.76 

1.77 
1.78 
1.79 

2.9241 
2.9584 
2.9929 

3.0276 
3.0625 
3.0976 

3.1329 
3.1684 
3.2041 

1.30767 
1.31149 
1.31529 

1.31909 
1.32288 
1.32665 

1.33041 
1.33417 
1.33791 

4.13521 
4.14729 
4.15933 

4.17133 

4.18330 
4.19524 

4.20714 
4.21900 
4.23084 

5.00021 
5.08845 
5.17772 

5.26802 
5.35938 
5.45178 

5.54523 
5.63975 
5.735.34 

1.19582 
1.19815 
1.20046 

1.20277 
1.20507 
1.20736 

1.20964 
1.21192 
1.21418 

2.57631 
2.58133 
2.58632 

2.59129 
2.59625 
2.60118 

2.60610 
2.61100 
2.61588 

5.55050 
5.56130 
5.57205 

5.582VV 
5.59344 
5.60408 

5.61467 
5.62523 
5.63574 

1.80 

3.2400 

1.34164 

4.24264 

5.83200 

1.21044 

2.62074 

5.64622 

1.81 
1.82 
1.83 

1.84 
1.85 
1.86 

1.87 
1.88 
1.89 

3.2761 
3.3124 
3.3489 

3.3856 
3.4225 
3.4596 

3.4969 
3.5344 
3.5721 

1.34536 
1.34907 
1.35277 

1.35647 
1.36015 
1.36382 

1.36748 
1.37113 
1.37477 

4.25441 
4.26615 

4.27785 

4.28952 
4.30116 
4.31277 

4.32435 
4.33590 
4.34741 

5.92974 
6.02857 
6.12849 

6.22950 
6.33162 
6.43486 

6.53920 
6.64467 
6.75127 

1.21809 
1.22093 
1.22316 

1.22539 
1.22760 
1.22981 

1.23201 
1.23420 
1.23639 

2.(32559 
2.03041 
2.63522 

2.64001 
2.64479 
2.64954 

2.65428 
2.65901 
2.66371 

5.656(35 
5.66705 
5.67741 

5.68773 
5.69802 

5.70827 

5.71848 
5.728(35 
5.73879 

1.90 

1.91 
1.92 
1.93 

1.94 
1.95 
1.96 

1.97 
1.98 
1.99 

3.6100 

1.37840 

4.35890 

6.85900 

1.23856 

2.66840 

5.74890 

3.6481 
3.6864 
3.7249 

3.7636 
3.8025 
3.8416 

3.8809 
3.9204 
3.9601 

1.38203 
1.38564 
1.38924 

1.39284 
1.39(342 
1.40000 

1.40357 
1.40712 
1.41067 

4.37035 
4.38178 
4.39318 

4.40454 

4.41588 
4.42719 

4.43847 
4.44972 
4.46094 

6.96787 
7.07789 
7.18906 

7.30138 
7.41488 
7.52954 

7.64537 
7.76239 
7.88060 

1.24073 
1.24289 
1.24.505 

1.24719 
1.24933 
1.25146 

1.25359 
1.25571 
1.25782 

2.67307 
2.67773 
2.68237 

2.68700 
2.69161 
2.69620 

2.70078 
2.70534 
2.70989 

5.75897 
5.76900 
5.77900 

5.78896 
5.79889 
5.80879 

5.81805 
5.82848 
5.83827 

U] 


Powers  and  Roots 


XI 


n 

W2 

Vn 

VlOn 

n^ 

^ 

^10  n 

^100  n 

2.00 

4.0000 

1.41421 

4.47214 

8.00000 

1.25992 

2.71442 

5.84804 

2.01 
2.02 
2.03 

2.04 
2.05 
2.06 

2.07 
2.08 
2.09 

4.0401 
4.0804 
4.1209 

4.1616 
4.2025 
4.2436 

4.2849 
4.3264 
4.3«)81 

1.41774 
1.42127 
1.42478 

1.42829 
1.43178 
1.43527 

1.43875 
1.44222 
1.44568 

4  48330 
4.49444 
4.50555 

4.51664 
4.52709 
4.53872 

4.54973 
4.5()070 
4.57165 

8.120(50 
8.24241 
8.36543 

8.48966 
8.(51512 
8.74182 

8.86974 
8.99891 
9.12933 

1.26202 
1.2(>111 
1.26619 

1.2()827 
1.27033 
1.27240 

1.27445 
1.27650 
1.27854 

2.71893 
2.72344 
2.72792 

2.73239 
2.73(585 
2.74129 

2.74572 
2.75014 
2.75454 

5.85777 
5.86746 
5.87713 

5.88677 
5.89637 
5.90594 

5.91548 
5.92499 
5.93447 

2.10 

4.4100 

1.44914 

4.58258 

9  26100 

1.28058 

2.75892 

5.94392 

2.11 
2.12 
2.13 

2.14 
2.15 
2.16 

2.17 

2.18 
2.19 

4.4521 
4.4944 
4.5369 

4.5796 
4.6225 
4.6656 

4.7089 
4.7524 
4.7961 

1.45258 
1.45602 
1.45945 

1.46287 
1.46629 
1.46969 

1.47309 
1.47648 
1.47986 

4.59347 
4.60435 
4.61519 

4.62601 
4.63681 
4.64758 

4.65833 
4.66905 
4.67974 

9.39393 
9.52813 
9.66360 

9.80034 
9.93838 
10.0777 

10.2183 
10.3602 
10.5035 

1.28261 
1.28463 
1.28665 

1.28866 
1.2906(5 
1.29266 

1.29465 
1.29664 
1.29862 

2.76330 
2.7(5766 
2.77200 

2.77633 
2.78065 
2.78495 

2.78924 
2.79352 
2.79779 

5.95334 
5.96273 
5.97209 

5.98142 
5.99073 
6.00000 

6.00925 
6.01846 
6.02765 

2.20 

4.8400 

1.48324 

4.69042 

10.6480 

1.30059 

2.80204 

6.03681 

2.21 
2.22 
2.23 

2.24 
2.2,5 
2.26 

2.27 
2.28 
2.29 

4.8841 
4.9284 
4.9729 

5.0176 
5.0625 
5.1076 

5.1529 
5.1984 
5.2441 

1.48661 
1.48997 
1.49332 

1.49666 
1.50000 
1.50333 

.1.5066.5 
1.50997 
1.51327 

4.70106 
4.71169 
4.72229 

4.73286 
4.74342 
4.75395 

4.76445 
4.77493 
4.78539 

10.7939 
10.9410 
11.0896 

11.2394 
11.3906 
11.5432 

11.6971 
11.8524 
12.0090 

1.30256 
1.30452 
1.30648 

1.30843 
1.31037 
1.31231 

1.31424 
1.31617 
1.31809 

2.80(528 
2.81050 
2.81472 

2.81892 
2.82311 
2.82728 

2.83145 
2.83560 
2.815974 

6.04594 
6.05505 
6.06413 

6.07318 
6.08220 
6.09120 

6.10017 
6.10911 
6.11803 

2.30 

5.2900 

1  51658 

4.79583 

12.1670 

1.32001 

2.84387 

6.12693 

2.31 
2.32 
2.33 

2.34 
2.35 
2.36 

2.37 
2.38 
2.39 

5.3361 
5.3824' 
5.4289 

5.475(> 
5.5225 
5.5696 

5.6169 
5.()(;44 
5.7121 

1.51987 
1.52315 
1.52643 

1.52971 
1.53297 
1.53623 

1.53948 
1.54272 
1.54596 

4.80625 
4.81664 
4.82701 

4.83735 
4.84768 
4.85798 

4.86826 
4.87852 
4.88876 

12.3264 
12.4872 
12.6493 

12.8129 
12.9779 
13.1443 

13.3121 
13.4813 
13.6519 

1.32192 
1.32382 
1.32572 

1.32761 
1.32950 
1.33139 

1.33326 
1.33514 
1.33700 

2.84798 
2.85209 
2.85618 

2.86026 
2.86433 
2.86838 

2.87243 
2.87646 
2.88049 

6.13579 
6.144(53 
6.15345 

6.16224 
6.17101 
6.17975 

6.18846 
6.19715 
6.20582 

2.40 

5.7600 

1.54919 

4.89898 

13.8240 

1.3.3887 

2.884.50 

6.21447 

2.41 
2.42 
2.43 

2.44 

2.45 
2.46 

2.47 
2.48 
2.49 

5.8081 
5.8564 
5.9019 

5.9536 
6.0025 
6.0516 

6.1009 
6.1504 
6.2001 

1.55242 
1.55563 

1.55885 

1.56205 
l..")(r)25 
1.56H44 

1.57162 
1.57480 
1.57797 

4.90918 
4.91935 
4.92950 

4.93964 
4.<^975 
4.95984 

4.96991 
4.97996 
4.98;)<>9 

13.9975 
14.1725 
14.3489 

14.5268 
14.70(51 
14.8869 

15.0692 
15.2530 
15.4382 

1.34072 
1.34257 
1.34442 

1.34626 
1.34810 
1.34993 

1.36176 
1.35358 
1.35540 

2.88850 
2.89249 
2.89647 

2.90044 
2.90439 
2.90834 

2.91227 
2.91620 
2.92011 

6.22308 
6.231(58 
6.24025 

6.24880 
6.25732 
6.26583 

6.27431 
6.28276 
6.29119 

Xll 


Powers  and  Roots 


[li 


n 

tt2 

V^ 

\/Wn 

n^ 

</n 

^10^ 

S/lOOn 

2.50 

6.2500 

1.58114 

5.00000 

15.6250 

i.a^>72i 

2.92402 

6.29961 

2.51 
2.52 
2.53 

2.54 
2.55 
2.56 

2.57 
2.58 
2.59 

6.3001 
6.3501 
6.4009 

6.4516 
6.5025 
6.5536 

6.6049 
6.6564 
6.7081 

1.58430 
1.58745 
1.59060 

1.59374 
1.59687 
1.60000 

1.60312 
1.60624 
1.60935 

5.00999 
5.01996 
5.02991 

5.03984 
5.04975 
5.05964 

5.06952 
5.07937 
5.08920 

15.8133 
16.0030 
16.1943 

16.3871 
16.5814 
16.7772 

16.9746 
17.1735 
17.3740 

l.a5902 
1.36082 
1.36262 

1.36441 
1.36620 
1.36798 

1.36976 
1.37153 
1.373.30 

2.92791 
2.93179 
2.93567 

2.93953 
2.94338 
2.94723 

2.95106 

2.95488 
2.95869 

6.30799 
6.31636 
6.32470 

6.33303 
6.34133 
6.34960 

6.35786 
6.36610 
6.37431 

2.60 

6.7600 

1.61245 

5.09902 

17.57<50 

1.. 37507 

2.9(5250 

6.38250 

2.61 
2.62 
2.63 

2.64 
2.65 
2.66 

2.67 
2.68 
2.69 

6.8121 
6.8(544 
6.9169 

6.9696 
7.0225 
7.0756 

7.1289 
7.1824 
7.2361 

1.61555 
1.61864 
1.62173 

1.62481 
1.62788 
1.63095 

1.63401 
1.63707 
1.64012 

5.10882 
5.11859 
5.128,35 

5.13809 
5.14782 
5.15752 

5.16720 
5.17687 
5.18652 

17.7796 
17.9847 
18.1914 

18.3997 
18.6096 
18.8211 

19.0342 
19.2488 
19.4651 

1.37683 
1.37859 
1.38034 

1.38208 
1.38383 
1.38557 

1.38730 
1.38903 
1.3<)076 

2.96629 
2.97007 
2.97385 

2.97761 
2.98137 
2.98511 

2.98885 
2.99257 
2.99629 

6.390(58 
6.39883 
6.40696 

6.41507 
6.42316 
6.43123 

6.43928 
6.44731 
6.45531 

2.70 

7.2900 

1.64317 

5.19(515 

19.(5830 

1 .39248 

3.00000 

6.46330 

2.71 

2.72 
2.73 

2.74 
2.75 
2.76 

2.77 
2.78 
2.79 

7.3441 
7.3984 
7.4529 

7.5076 

7.5625 
7.6176 

7.6729 
7.7284 
7.7841 

1.64621 
1.64924 
1.65227 

1.65529 
1.65831 
1.66132 

1.66433 
1.66733 
1.67033 

5.20577 
5.21536 
5.22494 

5.23450 
5.24404 
5.25357 

5.26308 
5.27257 
5.28205 

19.9025 
20.1236 
20.3464 

20.5708 
20.7969 
21.0246 

21.2539 
21.4850 
21.7176 

1.39419 
1.39591 
1.39761 

1.39932 
1.40102 
1.40272 

1.40441 
1.40610 
1.40778 

3.00370 
3.00739 
3.01107 

3.01474 
3.01841 
3.02206 

.3.02570 
3.02934 
3.03297 

6.47127 
6.47922 
6.48715 

6.49507 
6.5029<j 
6.51083 

6.51868 
6.52(552 
6.53434 

2.80 

7.8400 

1.67332 

5.29150 

21.9520 

1.40946 

3.03659 

6.54213 

2.81 
2.82 
2.83 

2.84 

2.85 
2.86 

2.87 
2.88 
2.89 

7.8961 
7.9524 
8.0089 

8.0656 
8.1225 
8.1796 

8.2369 
8.2944 
8.3521 

1.67631 
1.67929 
1.68226 

1.68523 
1.68819 
1.69115 

1.69411 
1.69706 
1.70000 

5.30094 
5.31037 
5.31977 

5.32917 
5.338M 
5.34790 

5.35724 
5.36656 
5.37587 

22.1880 
22.4258 
22.6652 

22.9003 
23.1491 
23.3937 

23.6399 
23.8879 
24.1376 

1.41114 

1.41281 
1.41448 

1.41614 
1.41780 
1.41946 

1.42111 
1.42276 
1.42440 

3.04020 
3;04380 
3.04740 

3.05098 
3.05456 
3.05813 

3.06169 
3.0(5524 
3.06878 

6.54991 
6.55767 
6.56541 

6.57314 
6.58084 
6.58853 

6.59620 
6.60385 
6.61149 

2.90 

8.4100 

1.70294 

5.38516 

24.3890 

1.42604 

3.07232 

6.61911 

2.91 
2.92 
2.93 

2.94 
2.95 
2.96 

2.97 
2.98 
2.99 

8.4681 
8.5264 
8.5849 

8.6436 
8.7025 
8.7616 

8.8209 
8.8804 
8.9401 

1.70587 
1.70880 
1.71172 

1.71464 
1.71756 
1.72047 

1.72337 
1.72627 
1.72916 

5.39444 
5.40370 
5.41295 

5.42218 
5.43139 
5.44059 

5.44977 
5.45894 
5.46809 

24.(5422 
24.8971 
25.1538 

25.4122 

25.6724 
25.9343 

26.1981 
26.4(536 
26.7309 

1.42768 
1.42931 
1.43094 

1.43257 
1.43419 
1.43581 

1.43743 
1.43904 
1.44065 

3.07584 
3.07936 
3.08287 

3.08638 
3.08987 
3.09336 

3.09684 
3.10031 
3.10378 

6.(52671 
6.63429 
6.64185 

6.64940 
6.65693 
6.66444 

6.67194 
6.67942 
6.68688 

11] 


Powers  and  Roots 


Xlll 


n 

n^ 

V^ 

VlOn 

W8 

i/a 

^On 

^00  n 

3.00 

9.0000 

1.73205 

5.47723 

27.0000 

1.44225 

3.10723 

6.69433 

3.01 
3.02 
3.03 

3.04 
3.05 
3.06 

3.07 
3.08 
3.09 

9.0601 
9.1204 
9.1809 

9.2416 
9.3025 
9.3636 

9.4249 
9.4864 
9.5481 

1.7.34iM 
1.73781 
1.74069 

1.74356 
1.74642 
1.74929 

1.75214 
1.75499 
1.75784 

5.486.-^ 
5.49545 
5.50454 

5.51.362 
5.52268 
5.53173 

5.54076 
5.54977 
5.55878 

27.2709 
27.5436 
27.8181 

28.0W5 
28.3726 
28.6526 

28.9344 
29.2181 
29.5036 

1.44385 
1.44545 
1.44704 

1.44863 
1.45022 
1.45180 

1.45338 
1.454% 
1.45653 

3.11068 
3.11412 
3.11756 

3.12098 
3.12440 
3.12781 

3.13121 
3.13401 
3.13800 

6.70176 
6.70917 
6.71657 

(5.72395 
6.73132 
6.73866 

6.74600 
6.75331 
6.7(5061 

3.10 

9.6100 

1.70068 

5.56776 

29.7910 

1.45810 

3.14138 

6.767iK) 

3.11 
3.12 
3.13 

3.14 
3.15 
3.16 

3.17 
3.18 
3.19 

9.6721 
9.7^44 
9.7969 

9.a596 
9.9225 

9.9856 

10.0489 
10.1124 
10.1761 

1.76352 
1.76635 
1.76918 

1.77200 
1.77482 
1.77764 

1.78045 
1.78326 
1.78606 

5.57674 
5.58570 
5.59464 

5.60357 
5.61249 
5.62139 

5.63028 
5.63915 
5.64801 

30.0802 
30.3713 
30.6643 

30.9591 
31.2559 
31.5545 

31.8550 
32.1574 
32.4618 

1.45f)67 
1.46123 
1.46279 

1.46434 
1.465i)0 
1.46745 

1.46899 
1.47054 
1.47208 

3.14475 
3.14812 
3.15148 

3.15483 
3.15818 
3.16152 

3.16485 
3.16817 
3.17149 

6.77517 
6.78242 
6.78966 

6.79(588 
6.80409 
6.81128 

6.81846 
6.82562 
6.83277 

3.20 

10.2400 

1.78885 

5.65685 

32.7680 

1.47361 

3.17480 

6.83i)90 

3.21 
3.22 
3.23 

3.24 
3.25 
3.26 

3.27 
3.28 
3.29 

3.30 

10.3041 
10.3684 
10.4329 

10.4976 
10.5625 
10.6276 

10.6929 
10.7584 
10.8241 

1.79165 
1.79444 
1.79722 

1.80000 
1.80278 
1.80555 

1.80831 
1.81108 
1.81384 

5.(56569 
5.(;7450 
5.68331 

5.69210 

5.70088 
5.701M54 

5.71839 
5.72713 

5.73585 

33.0762 
33.3862 
33.6983 

34.0122 
34.3281 
M.6460 

34.9658 
35.2876 
35.6113 

1.47515 
1.476(58 
1.47820 

1.47973 
1.48125 

1.48277 

1.48428 
1.48579 
1.48730 

3.17811 
3.18140 
3.18469 

3.18798 
3.19125 
3.19452 

3.19778 
3.20104 
3.20429 

6.84702 
6.85412 
6.86121 

6.86829 
6.87534 
6.88239 

6.88942 
6.89643 
6.90344 

10.8^)00 

1.81659 

5.744o6 

35.9370 

1.48881 

3.20753 

6.91042 

3.31 
3.32 
3.33 

3.34 
3.35 
3.36 

3.37 
3.38 
3.39 

10.9561 
11.0224 
11.0889 

11.1556 
11.2225 
11.2896 

11.3569 
11.4244 
11.4921 

1.819;M 
1.82209 
1.82483 

1.82757 
1.83030 
1.83303 

1.83576 
1.83848 
1.84120 

6.75326 
5.76194 
5.77062 

5.77927 
5.78792 
5.79655 

5.80517 
5.81378 
5.82237 

36.2647 
36.5<>44 
36.9260 

37.2597 
37.5954 
37.9331 

38.2728 
38.6145 
38.9582 

1.49031 
1.49181 
1.49330 

1.49480 
1.49(529 
1.49777 

1.49926 
1.50074 
1.50222 

3.21077 
3.21400 
3.21722 

3.22044 
3.22365 
3.22686 

3.2.3006 
3.23325 
3.23643 

6.91740 
6.92436 
6.93130 

6.93823 
6.94515 
6.95205 

6.95894 
6.96582 
6.97268 

3.40 

11.5(500 

1.84391 

5.83095 

39.3040 

1.50.369 

3.2.31X51 

6.97953 

3.41 
3.42 
3.43 

3.44 
3.45 
3.46 

3.47 
3.48 
3.49 

11.6281 
11.6964 
11.7649 

11.8336 
11.9025 
11.9716 

12.0409 
12.1104 
12.1801 

1.846<52 
1.84932 
1.85203 

1.85472 
1.8.-)742 
1.86011 

1.86279 
1.86548 
1.86815 

5.83952 
5.84808 
5.85662 

5.86515 
5.87367 
5.88218 

5.89067 
5.89915 
5.90762 

39.(5518 
40.0017 
40.3536 

40.7076 
41.0636 
41.4217 

41.7819 
42.1442 
42.5085 

1.50517 
1.50664 
1.50810 

1.50957 
1.51103 
1.51249 

1.51394 
1.51540 
1.51685 

3.24278 
3.24.195 
3.24911 

3.25227 
3.25.142 
3.25856 

3.26169 
3.26482 
3.26795 

6.98(537 
6.99319 
7.00000 

7.00680 
7.013,58 
7.02035 

7.02711 
7.03385 
7.04058 

XIV 


Powers  and  Roots 


ui 


n 

in? 

V^ 

VlO/i 

n^ 

^n 

^ron 

<J\mn 

3.50 

12.2500 

1.87083 

5.91608 

42.8750 

1.51829 

3.27107 

7.047.30 

3.51 
3.52 
3.53 

3.54 
3.55 
3.56 

3.57 
3.58 
3.59 

12.3201 
12.3904 
12.4609 

12.5316 
12.6025 
12.6736 

12.7449 
12.8164 

12.8881 

1.87350 
1.87617 
1.87883 

1.88149 
1.88414 
1.88680 

1.88944 
1.89209 
1.89473 

5.92453 
5.93296 
6.94138 

5.94979 
5.95819 
5.96657 

5.97495 
5.98331 
5.99166 

43.24;B6 
43.6142 
43.9870 

44..3619 
44.7389 
45.1180 

45.4993 
45.8827 
46.2683 

1.51974 
1.52118 
1.52262 

1.52406 
1.52549 
1.52692 

1.52835 
1.52978 
1.53120 

3.27418 
3.27729 
3.28039 

3.28.348 
3.28657 
3.28965 

3.29273 
3.29580 
3.29887 

7.05400 
7.06070 
7.06738 

7.07404 

7.08070 
7.08734 

7.09397 
7.10059 
7.10719 

3.60 

12.9600 

1.89737 

6.00000 

46.6560 

1.53262 

3.30193 

7.11379 

3.61 
3.62 
3.63 

3.64 
3.65 
3.66 

3.67 
3.68 
3.69 

13.0321 
13.1044 
13.1769 

13.2496 
13.3225 
13.3956 

13.4689 
13.5424 
13.6161 

1.90000 
1.90263 
1.90526 

1.90788 
1.91050 
1.91311 

1.91572 
1.91833 
1.920<M 

6.00833 
6.01664 
6.02495 

6.03324 
6.04152 
6.04979 

6.05803 
6.06030 
6.07454 

47.0459 
47.4379 
47.8321 

48.2285 
48.6271 
49.0279 

49.4309 
49.8360 
50.2434 

1.53404 
1.53545 
1.53686 

1.53827 
1.53968 
1.54109 

1.54249 
1.54389 
1.. 54529 

3.30498 
3.30803 
3.31107 

3.31411 
3.31714 
3.32017 

3.32319 
3.32621 
3.32922 

7.12037 
7.12694 
7.13349 

7.14004 

7.14(557 
7.15309 

7.15960 
7.16610 

7.17258 

3.70 

13.6900 

1.92354 

6.08276 

50.6530 

1.54668 

3.33222 

7.17905 

3.71 
3.72 
3.73 

3.74 
3.75 
3.76 

3.77 
3.78 
3.79 

13.7641 
13.8384 
13.9129 

13.9876 
14.0625 
14.1376 

14.2129 
14.2884 
14.3641 

1.92614 
1.92873 
1.93132 

1.93391 
1.93649 
1.93907 

1.94165 
1.94422 
1.94679 

6.09098 
6.09918 
6.10737 

6.11555 
6.12372 
6.13188 

6.14003 
6.14817 
6.15630 

51.0648 
51.4788 
51.8951 

52.3136 
52.7344 
53.1574 

53.5826 
54.0102 
54.4399 

1.54807 
1.54946 
1.55085 

1.55223 
1.55362 
1.55500 

1.55637 
1.55775 
1.55912 

3.33522 
3.33822 
3.34120 

3.34419 
3.34716 
3.35014 

3.35310 
3.35607 
3.35902 

7.18552 
7.19197 
7.19840 

7.20483 
7.21125 
7.21765 

7.22405 
7.23043 
7.23680 

3.80 

14.4400 

1.94936 

6.16441 

54.8720 

1.56049 

3.36198 

7.24316 

3.81 
3.82 
3.83 

3.84 
3.85 
3.86 

3.87 
3.88 
3.89 

14.5161 
14.5924 
14.6689 

14.745C 
14.8225 
14.8996 

14.9769 
15.0544 
15.1321 

1.95192 
1.95448 
1.95704 

1.95959 
1.96214 
1.96469 

1.96723 
1.96977 
1.97231 

6.17252 
6.18061 
6.18870 

6.19677 
6.20484 
6.21289 

6.22093 
6.22896 
6.23699 

55.3063 
55.7430 
56.1819 

56.6231 
57.0666 
57.5125 

57.9606 
58.4111 
58.8039 

1.56186 
1.56322 
1.56459 

1.56595 
1.56731 
1.56866 

1.57001 
1.57137 
1.57271 

3.36492 
3.36786 
3.37080 

3.37373 
3.37666 
3.37958 

3.38249 
3.38540 
3.38831 

7.24950 
7.25584 
7.26217 

7.26848 
7.27479 
7.28108 

7.28736 
7.29363 
7.29989 

3.90 

15.2100 

1.97484 

6.24500 

59.3190 

1.57406 

3.39121 

7.30614 

3.91 
3.92 
3.93 

3.94 
3.95 
3.96 

3.97 
3.98 
3.99 

15.2881 
15.3664 
15.4449 

15.52.36 
15.6025 
15.6816 

15.7609 
15.8404 
15.9201 

1.97737 
1.97990 
1.98242 

1.98494 
1.98746 
1.98997 

1.99249 
1.99499 
1.99750 

6.25300 
6.26099 
6.26897 

6.27694 
6.28490 
6.29285 

6.30079 
6.30872 
6.31664 

59.7765 
60.2363 
60.6985 

61.1630 
61.6299 
62.0991 

62.5708 
63.0448 
63.5212 

1.57541 
1.57675 
1.57809 

1.57942 
1.58076 
1.58209 

1.58342 
1..58475 
1.58608 

3.39411 
3.39700 
3.39988 

3.40277 
3.405(>4 
3.40851 

3.41138 
3.41424 
3.41710 

7.31238 
7.318(51 
7.32483 

7.33104 
7.33723 
7.34342 

7.34960 
7.35576 
7.36192 

H] 

Powers  and  Hoots 

XV 

n 

n2 

V^ 

VlOn 

n^ 

ifH 

^10  n 

</imn 

4.00 

16.0000 

2.00000 

6.32456 

64.0000 

1.58740 

3.41995 

7.36806 

4.01 
4.02 
4.03 

4.04 
4.05 
4.06 

4.07 
4.08 
4.09 

16.0801 
16.1604 
16.2409 

16.3216 
16.4025 
16.4836 

16.5649 
16.6464 
16.7281 

2.00250 
2.00499 
2.00749 

2.00<)98 
2.01246 
2.01494 

2.01742 
2.01990 
2.022:^7 

6.33246 
6.34035 
6.34823 

6.35610 
6.;5639() 
6.37181 

6.37966 
6.;38749 
6.39531 

64.4812 
04.9648 
65.4508 

65.9393 
6(5.4301 
66.9234 

67.4191 
67.9173 
68.4179 

1.58872 
1.61K)04 
1.59136 

1.59267 
1.593i)9 
1.59530 

1.59661 
1.59791 
1.59922 

3.42280 
3.425(54 
3.42S48 

3.43131 
3.43414 
3.43697 

3.43979 
3.442(50 
3.44541 

7.37420 
7.38032 
7.38644 

7.39254 
7.39864 
7.40472 

7.41080 
7.41686 
7.42291 

4.10 

16.8100 

2.02485 

6.40312 

68.9210 

1.60052 

3.44822 

7.42896 

4.11 
4.12 
4.13 

4.14 
4.15 
4.16 

4.17 
4.18 
4.19 

16.8921 
16.9744 
17.0509 

17.1396 
17.2225 
17.3056 

17.3889 
17.4724 
17.5561 

2.02731 
2.02978 
2.03224 

2.03470 
2.03715 
2.03961 

2.04206 
2.04450 
2.04695 

6.41093 
6.41872 
6.42651 

6.43428 
6.44205 
0.44981 

6.45755 
6.46529 
(5.47302 

69.4265 
69.9345 
70.4450 

70.9579 
71.4734 
71.9913 

72.5117 
73.0346 
73.5601 

1.60182 
1.60312 
1.60441 

1.60571 
1.60700 
1.60829 

1.60958 
1.61086 
1.61215 

3.45102 
3.45382 
3.45661 

3.45939 
3.4(5218 
3.46496 

3.46773 
3.47050 
3.47327 

7.4:^499 
7.44102 
7.44703 

7.45304 
7.45004 
7.46502 

7.47100 
7.47697 
7.48292 

4.20 

17.6400 

2.04939 

6.48074 

74.0880 

1.61343 

3.47603 

7.48887 

4.21 
4.22 
4.23 

4.24 
4.25 
4.26 

4.27 
4.28 
4.29 

17.7241 

17.8084 
17.8929 

17.9776 
18.0625 
18.1476 

18.2329 
18.3184 
18.4041 

2.05183 
2.05426 
2.05670 

2.05913 
2.0(3155 
2.06398 

2.06640 
2.0(3882 
2.07123 

6.48845 
6.49(315 
6.50384 

6.51153 
6.51920 
6.52687 

6.53452 
6.54217 
6.54981 

74.6185 
75.1514 
75.6870 

76.2250 
76.7(356 
77.3088 

77.8545 
78.4028 
78.9536 

1.61471 
1.61599 
1.61726 

1.61853 
1.61981 
1.62108 

1.62234 
1.62361 
1.62487 

3.47878 
3.48154 
3.48428 

3.48703 
3.48977 
3.49250 

3.49523 
3.49796 
3.50068 

7.49481 
7.50074 
7.50666 

7.51257 
7.51847 
7.52437 

7.53025 
7.5;5(512 
7.54199 

4.30 

18.4900 

2.07364 

6.55744 

79.5070 

1.62613 

3.50340 

7.547&4 

4.31 
4.32 
4.33 

4.34 
4.35 
4.36 

4.37 
4.38 
4.39 

18.5761 
18.()(J24 
18.7489 

18.8356 
18.9225 
19.0096 

19.0969 
19.1844 
19.2721 

2.07005 

2.07846 
2.08087 

2.08327 
2.08567 
2.08806 

2.09045 
2.092M 
2.09523 

6.56506 
6.57267 
6.58027 

6.58787 
6.59545 
6.60303 

6.61060 
6.61816 
6.62571 

80.0630 
80.6216 
81.1827 

81.7465 
82.3129 
82.8819 

83.4535 
84.0277 
84.6045 

1.62739 
1.62865 
1.62991 

1.63116 
1.63241 
1.63366 

1.63491 
1.63619 
1.63740 

3.50611 

3.50882 
3.51153 

3.51423 

3.51692 
3.51902 

3.52231 
3.52499 
3.52767 

7.55369 
7.5,5953 
7.56535 

7.57117 
7.576r8 
7.58279 

7.58858 
7.59436 
7.60014 

4.40 

19.3600 

2.09762 

6.63325 

85.1840 

1.63864 

3.530.35 

7.6a')90 

4.41 
4.42 
4.43 

4.44 
4.45 
4.46 

4.47 
4.48 
4.49 

19.4481 
19.5364 
19.6249 

19.7136 
19.8025 
19.8916 

19.9809 
20.0704 
20.1001 

2.10000 
2.10238 
2.10476 

2.10713 
2.10950 
2.11187 

2.11424 
2.11(360 
2.11896 

6.64078 
6.&i831 
6.65582 

6.66333 
(5.67083 
6.67832 

6.68581 
6.69.-528 
6.70075 

85.7661 
86.3509 
86.9383 

87.5284 
88.1211 
88.7165 

89.3146 
89.9154 
90.5188 

1.(53988 
1.64112 
1.64236 

1.64359 
1.64483 
1.64606 

1.64729 
1.64851 
1.64974 

3.53302 
3.53569 
3.53835 

3.54101 
3.54367 
3.54632 

3.54897 
3.55162 
3.55426 

7.61166 
7.61741 
7.62316 

7.62888 
7.6.">«51 
7.64032 

7.64603 
7.(55172 
7.65741 

xvi 

Vi 

3wers  and  Roots 

cn 

n 

n^ 

Vn 

VlOw, 

n^ 

^n 

^lOn 

^100 1* 

4.50 

20.2500 

2.12132 

6.70820 

91.1250 

1.65096 

3.55(389 

7.66309 

4.51 
4.52 
4.53 

4.54 
4.55 
4.56 

4.57 

4.58 
4.59 

20.3401 
20.4304 
20.5209 

20.6116 
20.7025 
20.7936 

20.8849 
20.9764 
21.0681 

2.12368 
2.12603 
2.12838 

2.13073 
2.13307 
2.13542 

2.13776 
2.14009 
2.14243 

6.71.565 
6.72309 
6.73053 

6.73795 
6.74537 
6.75278 

6.76018 
6.76757 
6.77495 

91.73.39 
92.34.54 
92.9597 

93.5767 
94.1964 
94.8188 

95.4440 
96.0719 
96.7026 

1.65219 
1.65.341 
1.654(32 

1.65584 
1.65706 
1.65827 

1.65948 
1.66069 
1.66190 

3.55953 
3.56215 
3.56478 

3.56740 
3.57002 
3.57263 

3.57524 
3.57785 
3.58045 

7.66877 
7.67443 
7.68009 

7.68573 
7.69137 
7.69700 

7.70262 
7.70824 
7.71384 

4.60 

21.1600 

2.14476 

6.78233 

97.3360 

1.66310 

3.58305 

7.71944 

4.61 
4.62 
4.63 

4.64 
4.65 
4.66 

4.67 
4.68 
4.69 

21.2521 
21.3444 
21.4369 

21.5296 
21.6225 
21.7156 

21.8089 
21.9024 
21.9961 

2.14709 
2.14942 
2.15174 

2.15407 
2.15639 
2.15870 

2.16102 
2.16333 
2.16564 

6.78970 
6.79706 
6.80441 

6.81175 
6.81909 
6.82(342 

6.83374 
6.84105 
6.848.% 

97.9722 
98.6111 
99.2528 

99.8973 
100,545 
101.195 

101.848 
102.503 
103.162 

1.66431 
1.66551 
1.66671 

1.66791 
1.66911 
1.67030 

1.67150 
1.072(39 

1.67388 

3.58564 
3.58823 
3.59082 

3.59,340 
3.59598 
3.59856 

3.60113 
3.60370 
3.60626 

7.72503 
7.73061 
7.73619 

7.74175 
7.74731 

7.75286 

7.75840 
7.76394 
7.76946 

4.70 

22.0900 

2.16795 

6.85565 

103.823 

1.67507 

3.60883 

7.77498 

4.71 

4.72 
4.73 

4.74 
4.75 
4.76 

4.77 
4.78 
4.79 

22.1841 

22.2784 
22.3729 

22.4676 
22.5625 
22.6576 

22.7529 

22.8484 
22.9441 

2.17025 
2.17256 
2.17486 

2.17715 
2.17945 
2.18174 

2.18403 

2.18(;;;2 

2.18861 

6.86294 
6.87023 
6.87750 

6.88477 
6.89202 
6.89928 

6.90652 
6.91375 
6.92098 

104.487 
105.154 
105.824 

106.496 
107.172 
107.850 

108.531 
109.215 
109.902 

1.67626 
1.67744 
1.67863 

1.67981 
1.(38099 
1.68217 

1.683.34 
1.68452 
1.68569 

3.61138 
3.61394 
3.61649 

3.61903 
3.02158 
3.62412 

3.62665 
3.62919 
3.63172 

7.78049 
7.78599 
7.79149 

7.79697 
7.80245 
7.80793 

7.81339 
7.81885 
7.82429 

4.80 

23.0400 

2.19089 

6.92820 

110.592 

1.68687 

3.63424 

7.82974 

4.81 

4.82 
4.83 

4.84 
4.85 
4.86 

4.87 
4.88 
4.89 

23.1361 
23.2324 
23.3289 

23.4256 
23.5225 
23.6196 

23.7169 
23.8144 
23.9121 

2.19317 
2.19545 
2.19773 

2.20000 
2.20227 
2.20454 

2.20681 
2.20907 
2.21133 

6.9.3542 
6.94262 
6.94982 

6.9.5701 
6.96419 
6.97137 

6.978.54 
6.98570 
6.99285 

111.285 
111.980 
112.679 

113..380 
114.084 
114.791 

115.501 
116.214 
116.930 

1.68804 
1.68920 
1.69037 

1.69154 
1.69270 
1.69386 

1.69503 
1.69619 
1.69734 

3.63676 
3.63928 
3.64180 

3.64431 

3.64682 
3.64932 

3.65182 
3.65432 
3.65681 

7.83517 
7.84059 
7.84601 

7.85142 
7.85683 
7.86222 

7.86761 

7.87299 
7.87837 

4.90 

24.0100 

2.21359 

7.00000 

117.649 

1.69850 

3.65931 

7.88374 

4.91 
4.92 
4.93 

4.94 
4.95 
4.96 

4.97 
4.98 
.4.99 

24.1081 
24.20(34 
24.3049 

24.4036 
24..5025 
24.6016 

24.7009 
24.8004 
24.9001 

2.21.585 
2.21811 
2.22036 

2.22261 
2.22486 
2.22711 

2.22935 
2.23159 
2.23383 

7.00714 
7.01427 
7.02140 

7.02851 
7.03562 
7.04273 

7.04982 
7.05691 
7.0(5399 

118.371 
119.095 
119.823 

120.554 
121.287 
122.024 

122.763 
123.506 
124.251 

1.69965 
1.70081 
1.701()6 

1.70311 
1.70426 
1.70.540 

1.70655 
1.70769 
1.70884 

3.66179 
3.66428 
3.66676 

3.66924 
3.67171 
3.67418 

3.67(365 
3.67911 
3.68157 

7.88909 
7.89445 
7.89979 

7.90513 
7.91046 
7.91578 

7.92110 
7.92641 
7.93171 

UJ 


Powers  and  Roots 


XVll 


n 

W2 

VTi 

VlOn 

n^ 

^ 

^10  w 

^100  n 

5.00 

25.0000 

2.23607 

7.07107 

125.000 

1.70993 

3.68403 

7.93701 

5.01 
5.02 
5.03 

5.04 
5.05 
5.06 

5.07 
5.08 
5.09 

25.1001 
25.2004 
25.3009 

25.4016 
25.5025 
25.6036 

25.7049 
25.8064 
25.9081 

2.23830 
2.24054 
2.24277 

2.24499 
2.24722 
2.24944 

2.25167 
2.25389 
2.25610 

7.07814 
7.08520 
7.09225 

7.09930 
7.10654 
7.11337 

7.12039 
7.12741 
7.13442 

125.752 
12().50() 
127.264 

128.024 
128.788 
129.554 

130.324 
131.097 
131.872 

1.71112 
1.71225 
1.71339 

1.71452 
1.715(56 
1.71679 

1.71792 
1.71i)05 
1.72017 

3.68649 
3.68894 
3.69138 

3.69383 
3.69627 
3.69871 

3.70114 
3.70:357 
3.70600 

7.*)4229 
7.94757 
7.95285 

7.95811 
7.96337 
7.96863 

7.97387 
7.97911 
7.984)34 

6.10 

26.0100 

2.25832 

7.14143 

132.051 

1.72i;^ 

3.70843 

7.98957 

5.11 
5.12 
5.13 

5.14 
5.15 
5.16 

5.17 
5.18 
5.19 

26.1121 
26.2144 
26.3169 

26.4196 
26.5225 
26.6256 

26.7289 
26.8324 
26.9361 

2.26053 
2.20274 
2.26495 

2.26716 
2.26936 
2.27156 

2.27376 
2.275<)() 
2.27816 

7.14843 
7.15542 
7.16240 

7.16938 
7.17635 
7.18;>3l 

7.19027 
7.19722 
7.20417 

133.433 
134.218 
135.006 

135.797 
13(5.591 
137.388 

138.188 
138.992 
139.798 

1.72242 
1.72;i55 
1.724(57 

1.72579 
1.72691 
1.72802 

1.72914 
1.73025 
1.73137 

3.71085 
3.71327 
3.71569 

3.71810 
3.72051 
3.72292 

3.72532 
372772 
3.73012 

7.9^)479 
8.00000 
8.00520 

8.01040 
8.01559 
8.02078 

8.02596 
8.03113 
8.03629 

5.20 

27.0400 

2.28035 

7.21110 

140.608 

1.73248 

3.73251 

8.04145 

5.21 
5.22 
5.23 

5.24 
5.25 
5.26 

5.27 
5.28 
5.29 

27.1441 
27.2484 
27.3529 

27.4576 
27.5(525 
27.6676 

27.7729 

27.8784 
27.9841 

2.28254 
2.28473 
2.28692 

2.28910 
2.29129 
2.29a47 

2.29565 
2.29783 
2.30000 

7.21803 
7.22496 
7.23187 

7.23878 
7.24569 
7.25259 

7.25<)48 

i.mm 

7.27324 

141.421 
142.237 
143.056 

143.878 
144.703 
145.532 

146.363 
147.198 
148.036 

1.73359 
1.73470 

1.73580 

1.73(591 
1.73801 
1.73912 

1.74022 
1.74132 
1.74242 

3.73490 
3.73729 
3.73968 

3.74206 
3.74443 
3.74681 

3.74918 
3.75155 
3.75392 

8.04(560 
8.05175 
8.05689 

8.06202 
8.06714 
8.07226 

8.07737 
8.08248 
8.08758 

5.30 

28.0900 

2.30217 

7.28011 

148.877 

1.74351 

3.75629 

8.0i)267 

5.31 
5.32 
5.33 

5.34 
5.35 
5.36 

5.37 
5.38 
5.39 

28.1CJ61 
28.:3024 
28.4089  ■ 

28.5156 
28.6225 
28.729«j 

28.8369 
28.9444 
29.0521 

2.30434 
2.30651 
2.30868 

2.31084 
2.31301 
2.31517 

2.31733 
2.31948 
2.32164 

7.28<;97 
7.29;i83 
7.30068 

7.30753 
7.31437 
7.32120 

7.32803 
7.33485 
7.34166 

149.721 
loO.odO 
151.419 

152.273 
1.53.130 
153.991 

154.854 
155.721 
156.591 

1.74461 
1.74570 
1.74680 

1.74789 
1.748i)8 
1.75007 

1.75116 
1.75224 
1.75333 

3.75865 
3.76101 
3.76336 

3.76571 
3.76806 
3.77041 

3.77275 
3.77609 
3.77743 

8.09776 
8.10284 
8.10791 

8.11298 
8.11804 
8.12310 

8.12814 
8.13319 
8.13822 

5.40 

29.1600 

2.32379 

7.34847 

157.4<)4 

1.75441 

3.77976 

8.14325 

5.41 
5.42 
5.43 

5.44 
5.45 
5.46 

5.47 
5.48 
5.49 

29.2681 
29.3764 
29.4849 

29.5936 
29.7025 
29.8116 

29.9209 
;i0.O304 
30.1401 

2.325iH 
2.3280i) 
2.33024 

2.33238 
2.33452 
2.33666 

2.33880 
2.34094 
2.34307 

7.35.527 
7.36206 
7.3688.5 

7.37.5(>4 
7.38241 
7.38918 

7.39594 
7.40270 
7.40945 

158.340 
159.220 
160.103 

160.989 
161.879 
162.771 

163  667 
164.5()7 
165.469 

1.75549 
1.75657 
1.75765 

1.75873 
1.75981 
1.76088 

1.76196 
1.7(5303 
1.76110 

3.78209 
3.78442 
3.78675 

3.78907 
3.79139 
3.79371 

3.79603 
3.79834 
3.80065 

8.14828 
8.15329 
8.15831 

8.16331 
8.16831 
8.17330 

8.17829 
8.18327 
8.18824 

xviii 

Powers  and  Roots 

Lli 

n 

n2 

Vn 

\/Wn 

n^ 

</n 

^10  n 

s/lOOw, 
8.19321 

6.60 

30.2500 

2.34521 

7.41620 

166.375 

1.76517 

3.80295 

5.51 
5.52 
5.53 

5.54 
5.55 
5.56 

5.57 
5.58 
5.59 

30.3601 
30.4704 
30.5809 

30.6916 
30.8025 
30.9136 

31.0249 
31.1364 
31.2481 

2.34734 
2.34947 
2.35160 

2.35372 
2.35584 
2.35797 

2.36008 
2.36220 
2.36432 

7.42294 
7.42967 
7.43640 

7.44312 
7.44983 
7.45654 

7.46324 
7.46994 
7.47663 

167.284 
168.197 
169.112 

170.031 
170.954 
171.880 

172.809 
173.741 
174.677 

1.76624 
1.76731 
1.76838 

1.76944 
1.77051 
1.77157 

1.77263 
1.77369 
1.77475 

3.80526 
3.80756 
3.80985 

3.81215 
3.81444 
3.81673 

3.81902 
3.82130 
3.82358 

8.19818 
8.20313 
8.20808 

8.21303 
8.21797 
8.22290 

8.22783 
8.23275 
8.23766 

6.60 

31.3600 

2.36643 

7.48331 

175.616 

1.77581 

3.82586 

8.24257 

5.61 
5.62 
5.63 

5.64 
5.65 
5.66 

5.67 
5.68 
5.69 

31.4721 
31.5844 
31.6969 

31.8096 
31.9225 
32.0356 

32.1489 
32.2624 
32.3761 

2.36854 
2.37065 
2.37276 

2.37487 
2.37697 
2.37908 

2.38118 
2.38328 
2.38537 

7.48999 
7.49667 
7.50333 

7.50999 
7.51665 
7.52330 

7.52994 
7.53658 
7.54321 

176.558 
177.504 
178.454 

179.406 
180.362 
181.321 

182.284 
183.250 
184.220 

1.77686 
1.77792 
1.77897 

1.78003 
1.78108 
1.78213 

1.78318 
1.78422 
1.78527 

3.82814 
3.83041 
3.83268 

3.83495 
3.83722 
3.83948 

3.84174 
3.84399 
3.84625 

8.24747 
8.25237 
8.25726 

8.26215 
8.26703 
8.27190 

8.27677 
8.28164 
8.28649 

5.70 

32.4900 

2.38747 

7.54983 

185.193 

1.78632 

3.84850 

8.29134 

5.71 
5.72 
5.73 

5.74 
5.75 
5.76 

5.77 
5.78 
5.79 

32.6041 
32.7184 
32.8329 

32.9476 
33.0625 
33.1776 

33.2929 
33.4084 
33.5241 

2.38956 
2.39165 
2.39374 

2.39583 
2.39792 
2.40000 

2.40208 
2.40416 
2.40624 

7.55645 
7.56307 
7.56968 

7.57628 
7.58288 
7.58947 

7.59605 
7.60263 
7.60920 

186.169 
187.149 
188.133 

189.119 
190.109 
191.103 

192.100 
193.101 
194.105 

1.78736 
1.78840 
1.78944 

1.79048 
1.79152 
1.79256 

1.79360 
1.79463 
1.79567 

3.85075 
3.85300 
3.85524 

3.85748 
3.85972 
3.86196 

3.86419 
3.86642 
3.86865 

8.2^)619 
8.30103 
8.30587 

8.31069 
8.31552 
8.32034 

8.32515 
8.32995 
8.33476 

6.80 

33.6400 

2.40832 

7.61577 

195.112 

1.79670 

3.87088 

8.33955 

5.81 
5.82 
5.83 

5.84 
5.85 
5.86 

5.87 
5.88 
5.89 

33.7561 
33.8724 
33.9889 

34.1056 
34.2225 
34.3396 

34.4569 
34.5744 
34.6921 

2.41039 
2.41247 
2.41454 

2.41661 
2.41868 
2.42074 

2.42281 
2.42487 
2.42693 

7.62234 
7.62889 
7.63544 

7.64199 
7.64853 
7.65506 

7.66159 

7.66812 
7.67463 

196.123 
197.137 
198.155 

199.177 
200.202 
201.230 

202.262 
203.297 
204.336 

1.79773 
1.79876 
1.79979 

1.80082 
1.80185 
1.80288 

1.80390 
1.80492 
1.80595 

3.87310 
3.87532 
•3.87754 

3.87975 
3.88197 
3.88418 

3.88639 
3.88859 
3.89080 

8.34134 
8.34913 
8.35390 

8.35868 
8.36345 
8.36821 

8.37297 
8.37772 
8.38247 

6.90 

34.8100 

2.42899 

7.68115 

205.379 

1.80697 

3.89P.00 

8.38721 

5.91 
5.92 
5.93 

5.94 
5.95 
5.96 

5.97 
5.98 
5.99 

34.9281 
35.0464 
35.1649 

35.2836 
35.4025 
35.5216 

a5.6409 
35.7604 
35.8801 

2.43105 
2.43311 
2.43516 

2.43721 
2.43926 
2.44131 

2.44336 
2.44540 

2.44745 

7.68765 
7.69415 
7.70065 

7.70714 
7.71362 
7.72010 

7.72658 
7.73305 
7.73951 

206.425 
207.475 
208.528 

209.585 
210.645 
211.709 

212.776 
213.847 
214.922 

1.80799 
1.80901 
1.81003 

1.81104 
1.81206 
1.81307 

1.81409 
1.81510 
1.81611 

3.89519 
3.89739 
3.89958 

3.90177 
3.90396 
3.90615 

3.90833 
3.91051 
3.91269 

8.39194 
8.39667 
8.40140 

8.40612 
8  41083 
8.41554 

8.42025 
8.42494 
8.42964 

n] 


Powers  and  Roots 


n 

n2 

VE 

VlOn 

»8 

^ 

^S^IOn 

VHiJn 

6.00 

36.0000 

2.44949 

7.74597 

216.000 

1.81712 

3.91487 

8.43-1:33 

6.01 
6.02 
6.03 

6.04 
6.05 
6.06 

6.07 
6.08 
6.09 

36.1201 
36.2404 
36.3609 

36.4816 
S().602.5 
36.7236 

36.8449 
36.96(;4 
37.0881 

2.45153 
2.45357 
2.45561 

2.45764 
2.459(37 
2.46171 

2.46374 
2.46577 
2.46779 

7.75242 
7.75887 
7.76531 

7.77174 
7.77817 
7.78460 

7.79102 
7.79744 

7.80;385 

217.082 
218.167 
219.256 

220.349 
221.445 
222.545 

223.649 
224.756 

225.8(57 

1.81813 
1.81914 
1.82014 

1.82115 
1.82215 
1.82316 

1.82416 
1.82516 
1.82616 

3.91704 
3.91921 
3.92138 

3.92.3.'>5 

3.92787 

3.9,'5003 
3.93219 
3.9:34:34 

8.43901 
8.44:3(59 
8.44836 

8.45:303 
8.457(59 
8.46235 

8.46700 
8.47165 
8.47(529 

6.10 

37.2100 

2.46982 

7.81025 

226.981 

1.82716 

3.93650 

8.48093 

6.11 
6.12 
6.13 

6.14 
6.15 
6.16 

6.17 
6.18 
6.19 

37.3(?21 
37.4544 
37.5769 

37.69<>6 
37.8225 
37.i^56 

38.0689 
38.1924 
38.3161 

2.47184 
2.47386 
2.47588 

2.47790 
2.479<)2 
2.48193 

2.48395 
2.485% 
2.48797 

7.81665 
7.82304 
7.82943 

7.83582 
7.84219 
7.84857 

7.85493 
7.86130 
7.86766 

228.099 
229.221 
230.346 

231.476 
232.()08 
233.745 

234.885 
230.029 
237.177 

1.82816 
1.82915 
1.83015 

1.83115 
1.83214 
1.83313 

1.83412 
1.83511 
1.83610 

3.93865 
3.94079 
3.94294 

3.94508 
3.94722 
3.94936 

3.95150 
3.95363 
3.95576 

8.4855G 
8.49018 
8.49481 

8.49942 
8.50403 
8.50864 

8.51324 
8.51784 
8.52243 

6.20 

38.4400 

2.48998 

7.87401 

238.328 

1.83709 

3.95789 

8.52702 

6.21 
6.22 
6.23 

6.24 
6.25 
6.26 

6.27 
6.28 
6.29 

38.5(i41 
38.6884 
38.8129 

38.9376 
39.062.5 
39.1876 

39.3129 
39.4384 
39.5(^1 

2.49199 
2.49.399 
2.4^)600 

2.49800 
2.50000 
2.50200 

2.50400 
2.50599 
2.50799 

7.88036 
7.88670 
7.89303 

7.89937 
7.90569 
7.91202 

7.91833 
7.92465 
7.93095 

239.483 
240.642 
241.804 

242.971 
244.141 
245.314 

246.492 
247.673 
248.858 

1.83808 
1.83906 
1.84005 

1.84103 

1.84202 
1.84300 

1.84398 
1.844% 
1. 84504 

3.9(5002 
3.9(5214 
3.96427 

3.96638 
3.96850 
3.97062 

3.97273 
3.97484 
3.97695 

8.5:3160 
8.5:3618 
8.54075 

8.54532 
8.54988 
8.55444 

8.55899 
8..56a54 
8.56808 

6.30 

39.6900 

2mm 

7.93725 

250.047 

1.84691 

3.97906 

8.57262 

6.31 
6.32 
6.33 

6.34 
6.35 
6.36 

6.37 
6.38 
6.39 

39.8161 
39.9424 
40.0689 

40.1956 
40.3225 
40.4496 

40.5769 
40.7044 
40.8321 

2.51197 
2.51396 
2.51595 

2.51794 
2.5]9<>2 
2.52190 

2.52389 
2.52587 
2.52784 

7.94355 
7.iM984 
7.95613 

7.96241 

7.96869 
7.97496 

7.98123 
7.98749 
7.99375 

251.240 
252.436 
253.636 

254.840 
2rKi.048 
257.259 

258.475 
259.6<)4 
2()0.917 

1.84789 
1.84887 
1.84984 

1.85082 
1.85179 
1.85276 

1.85373 
1.85470 
1.85567 

3.98116 
3.98326 
3.98536 

3.98746 
3.98956 
3.99165 

3.99374 
3.99583 
3.99792 

8.57715 
8.58168 
8.58620 

8.59072 
8.59524 
8.59975 

8.60425 
8.60875 
8.61.325 

6.40 

40.9600 

2.52<)82 

8.00000 

262.144 

1.85664 

4.00000 

8.61774 

6.41 
6.42 
6.43 

6.44 
6.45 
6.46 

6.47 
6.48 
6.49 

41.0881 
41.2164 
41.3449 

41.4736 
41.(K)25 
41.7316 

41.8609 
41.9904 
42.1201 

2.53180 
2.53377 
2.53574 

2.53772 
2.531^39 
2.54165 

2.54362 
2.54558 
2.54755 

8.00«)25 
8.01249 
8.01873 

8.02496 
8.03119 
8.03741 

8.04363 
8.04984 
8.05605 

263..375 
264.609 
265.848 

267.090 
268.3.36 
2(59.586 

270.840 
272.098 
273.359 

1.&5760 
1.85857 
1.85953 

1.86050 
1.86146 
1.86242 

1.86338 
1.86434 
1.86530 

4.00208 
4.00416 
4.00624 

4.008,32 
4.010:39 
4.01246 

4.01453 
4.01660 
4.01866 

8.62222 
8.(52671 
8.63118 

8.6.3.566 
8.64012 
8.64459 

8.64904 
8.(35350 
8.65795 

XX 

Powers  and  Roots 

[n 

n 

W,2 

yJn 

VlO»i 

n^ 

</n 

^10  w, 

s/lOOn 

6.50 

42.2500 

2.54951 

8.06226 

274.625 

1.86626 

4.02073 

8.06239 

6.51 
6.52 
6.53 

6.54 
6.55 
6.56 

6.57 
6.58 
6.59 

42.3801 
42.5104 
42.6409 

42.7716 
42.9025 
43.0336 

43.1649 
43.2964 
43.4281 

2.55147 
2.55343 
2.55539 

2.55734 
2.55930 
2.56125 

2.56320 
2.56515 
2.56710 

8.06846 
8.07465 
8.08084 

8.08703 
8.09321 
8.09938 

8.10555 
8.11172 
8.11788 

275.894 
277.168 
278.445 

279.726 
281.011 
282.300 

283.593 
284.890 
286.191 

1.86721 

1.86817 
1.86912 

1.87008 
1.87103 
1.87198 

1.87293 

1.87388 
1.87483 

4.02279 
4.02485 
4.02690 

4.02896 
4.03101 
4.03306 

4.03511 
4.03715 
4.03920 

8.()()683 
8.67127 
8.67570 

8.68012 
8.68455 
8.68896 

8.69338 
8.69778 
8.70219 

6.60 

43.5600 

2.56905 

8.12404 

287.496 

1.87578 

4.04124 

8.70659 

6.61 
6.62 
6.63 

6.64 
6.65 
6.66 

6.67 
6.68 
6.69 

43.6921 
43.8244 
43.9569 

44.0896 
44.2225 
44.3556 

44.4889 
44.6224 
44.7561 

2.57099 
2.57294 
2.57488 

2.57682 
2.57876 
2.58070 

2.58263 
2.58457 
2.58650 

8.13019 
8.13634 
8.14248 

8.14862 
8.15475 
8.16088 

8.16701 
8.17313 
8.17924 

288.805 
290.118 
291.434 

292.755 
294.080 
295.408 

296.741 
298.078 
2t)9.418 

1.87672 
1.87767 
1.87862 

1.87956 
1.88050 
1.88144 

1.88239 
1.88333 

1.88427 

4.04328 
4.04532 
4.04735 

4.049.39 
4.05142 
4.05345 

4.05548 
4.05750 
4.05953 

8.71098 
8.71537 
8.71976 

8.72414 

8.72852 
8.73289 

8.73726 
8.74162 
8.74598 

6.70 

44.8900 

2,58844 

8.18535 

300.763 

1.88520 

4.06155 

8.75034 

6.71 
6.72 
6.73 

6.74 
6.75 
6.76 

6.77 
6.78 
6.79 

45.0241 
45.1584 
45.2929 

45.4276 
45.5625 
45.6976 

45.8329 
45.9684 
46.1041 

2.59037 
2.59230 
2.59422 

2.59615 
2.59808 
2.60000 

2.60192 
2.60384 
2.60576 

8.19146 
8.19756 
8.20366 

8.20975 
8.21584 
8.22192 

8.22800 
8.23408 
8.24015 

302.112 
303.464 
304.821 

306.182 
307.547 
308.916 

310.289 
311.666 
313.047 

1.88G14 
1.88708 
1.88801 

1.88895 
1.88988 
1.89081 

1.89175 
1.89268 
1.89361 

4.06357 
4.06559 
4.06760 

4.06961 
4.07163 
4.07364 

4.07564 
4.07765 
4.07965 

8.75469 
8.75904 
8.76338 

8.76772 
8.77205 
8.77638 

8.78071 
8.78503 
8.78935 

6.80 

46.2400 

2.60768 

8.24621 

314.432 

1.89454 

4.08166 

8.79366 

6.81 
6.82 
6.83 

6.84 
6.85 
6.86 

6.87 
6.88 
6.89 

46.3761 
46.5124 
46.6489 

46.7856 
46.9225 
47.0596 

47.1969 
47.3344 
47.4721 

2.60960 
2.61151 
2.61343 

2.61534 
2.61725 
2.61916 

2.62107 
2.62298 
2.62488 

8.25227 
8.25833 
8.26438 

8.27043 
8.27647 
8.28251 

8.28855 
8.29458 
8.30060 

315.821 
317.215 
318.612 

320.014 
321.419 
322.829 

324.243 
325.661 
327.083 

1.89546 
1.89639 
1.89732 

1.89824 
1.89917 
1.90009 

1.90102 
1.90194 
l.i)0286 

4.08365 
4.08565 
4.08765 

4.08964 
4.09163 
4.09362 

4.09561 
4.09760 
4.09958 

8.79797 
8.80227 
8.80657 

8.81087 
8.81516 
8.81945 

8.82373 
8.82801 
8.83228 

6.90 

47.6100 

2.62679 

8.30662 

328.509 

1.90378 

4.10157 

8.83f)56 

6.91 
6.92 
6.93 

6.94 
6.95 
6.96 

6.97 

6.98 
6.99 

47.7481 
47.8864 
48.0249 

48.1636 
48..3025 
48.4416 

48.5809 
48.7204 
48.8601 

2.62869 
2.63059 
2.63249 

2.63439 
2.63629 
2.63818 

2.64008 
2.64197 
2.64386 

8.31264 
8.31865 
8.32466 

8.33067 
8.33667 
8.34266 

8.34865 
8.35464 
8.36062 

329.939 
331.374 
332.813 

334.255 
335.702 
337.154 

338.609 
340.068 
341.532 

1.90470 
1.90562 
1.90653 

1.90745 
1.90837 
1.90928 

1.91019 
1.91111 
1.91202 

4.10355 
4.10552 
4.10750 

4.10948 
4.11145 
4.11342 

4.11539 
4.11736 
4.11932 

8.84082 
8.84509 
8.84934 

8.85360 
8.85785 
8.86210 

8.86634 
8.87058 
8.87481 

II] 

Powers  and  Roots 

xxi 

n 

n2 

v^ 

VlOn 

W8 

</ii 

</10n 

^100  n 

7.00 

49.0000 

2.64576 

8.36660 

343.000 

1.912a3 

4.12129 

8.87904 

7.01 
7.02 
7.03 

7.04 
7.05 
7.06 

7.07 
7.08 
7.09 

49.1401 
49.2804 
49.4209 

49..T>16 
49.7025 
49.8436 

49.9849 
50.1264 
50.2681 

2.64764 
2.(>4963 
2.65141 

2.65330 
2.05518 
2.65707 

2.66895 
2.66083 
2.66271 

8.37257 
8.37854 
8.3^451 

8.39047 
8.39()43 
8.40238 

8.40833 
8.41427 
8.42021 

344.472 
345.948 
347.429 

348.914 
350.403 
351.896 

353.393 
354.895 
.3.5<).401 

1.91384 
1.91475 
1.91566 

1.916,57 
1.91747 
1.91838 

1.91929 
1.92019 
l.t)210{) 

4.12325 
4.12521 
4.12716 

4.12912 
4.13107 
4.13303 

4.13498 
4.13693 

4.1.3887 

8.88.327 
8.88749 
8.89171 

8.89592 
8.90013 
8.90434 

8.90854 
8.91274 
8.91(593 

7.10 

50.4100 

2.66458 

8.42615 

357.911 

1.92200 

4.14()S2 

8.92112 

7.11 
7.12 
7.13 

7.14 
7.15 
7.16 

7.17 
7.18 
7.19 

50.5521 
50.6944 
50.8369 

50.9796 
51.1225 
51.2656 

51.4089 
51. .5524 
51.6^1 

2.66646 
2.66833 
2.67021 

2.67208 
2.67395 
2.67582 

2.67769 
2.67955 
2.68142 

8.43208 
8.43801 
8.44393 

8.44985 
8.45577 
8.46168 

8.46759 
8.47.'U9 
8.47939 

359.425 
360.944 
362.467 

363.994 
365.526 
367.062 

368.602 
370.146 
371.695 

1.92290 
1.92380 
1.92470 

1.92560 
1.92650 
1.92740 

1.92829 
1.92919 
1.9,3008 

4.14276 
4.1-1470 
4.146(54 

4.14858 
4.15052 
4.15245 

4.15438 
4.15(531 
4.15824 

8.92531 
8.92949 
8.93367 

8.93784 
8.94201 
8.94618 

8.95034 
8.95450 
8.95866 

7.20 

51.8400 

2.68328 

8.48,528 

373.248 

1.93098 

4.1(5017 

8.96281 

7.21 
7.22 
7.23 

7.24 
7.25 
7.26 

7.27 
7.28 
7.29 

51.9841 
52.1284 
52.2729 

52.4176 
52.5()25 
52.7076 

62.8529 
52.9984 
53.1441 

2.68514 
2.68701 
2.68887 

2.69072 
2.69258 
2.69444 

2.69629 
2.(59815 
2.70000 

8.49117 
8.49706 
8.50294 

8.50882 
8.51469 
8.52056 

8.52643 
8.53229 
8.53816 

374.806 
37(5.367 
377.933 

379.503 
381.078 
382.657 

384.241 
385.828 
.387.420 

1.93187 
1.93277 
1.93366 

1.93456 
1.9:3544 
1.93633 

1.93722 
1.93810 
1.93899 

4.1(5209 
4.10402 
4.16594 

4.16786 
4.16978 
4.17169 

4.17361 
4.17552 
4.17743 

8.96696 
8.97110 
8.97524 

8.97938 
8.98,351 
8.98764 

8.99176 
8.99588 
9.00000 

7.30 

53.2900 

2.70185 

8.54400 

389.017 

1.93988 

4.17934 

9.00411 

7.31 
7.32 
7.33 

7.34 
7.35 
7.36 

7.37 
7.39 

53.43()1 
53.5824 
53.7289 

53.8756 
54.0225 
64.1696 

54..3169 
54.4(>44 
54.6121 

2.70370 
2.70.555 
2.70740 

2.70924 
2.71109 
2.71293 

2.71477 
2.71662 
2.7184(> 

8.54985 
8.55.570 
8.56154 

8.56738 
8.57.321 
8.57904 

8.58487 
8.59069 
8.59661 

390.618 
392.223 
393.833 

395.447 
397.0(;5 
398.688 

400..316 
401.947 
403.583 

1.94076 
1.94165 
1.94253 

1.94341 
1.944.30 
1.94518 

1.94606 
1.94694 
1.94782 

4.18125 
4.18315 
4.18506 

4.18696 
4.18886 
4.19076 

4.19266 
4.194.55 
4.19644 

9.00822 
9.01233 
9.01643 

9.02053 
9.02462 
9.02871 

9.03280 
9.0.36H9 
9.04097 

7.40 

54.7600 

2.72029 

8.602.33 

405.224 

1.94870 

4.198,34 

9.04504 

7.41 
7.42 
7.43 

7.44 
7.45 
7.46 

7.47 
7.48 
7.49 

64.9081 
55.0564 
56.2049 

55.:«36 
55.5025 
55.651() 

55.8009 
5;"). 9504 
56.1001 

2.72213 
2.72397 
2.72580 

2.72764 
2.72947 
2.73130 

2.7a313 
2.73496 
2.73679 

8.60814 
8.61394 
8.61974 

8.62554 
8.63134 
8.63713 

8.64292 
8.64870 
8.6544H 

406.869 
408.618 
410.172 

411.8,31 
413.494 
415.161 

416.833 
418.609 
420.190 

1.94957 
1.95045 
1.95132 

1.95220 
1.95307 
1.95.395 

1.95482 
1.955(59 
1.956,56 

4.20023 
4.20212 
4.20400 

4.20589 
4.20777 
4.20966 

4.21153 
4.21341 
4.21529 

9.04911 
9.05318 
9.05725 

9.06131 
9.06537 
9.06942 

9.07347 
9.07752 
9.08156 

xxu 

Powers  and  Ro 

ols 

Cn 

n 

n^ 

V^ 

VlOw 

n^ 

</n 

</l(in 

S/lOO/i 

7.60 

56.2500 

2.73861 

8.66025 

421.875 

1.95743 

4.21716 

9.08560 

7.51 
7.52 
7.53 

7.54 
7.55 
7.56 

7.57 
7.58 
7.59 

56.4001 
56.5504 
56.7009 

56.8516 
57.0025 
57.1536 

57.3049 
57.4564 
57.6081 

2.74044 
2.74226 

2.74408 

2.74591 
2.74773 
2.74955 

2.75136 
2.75318 
2.75500 

8.66603 
8.67179 
8.67756 

8.68332 
8.68907 
8.69483 

8.70057 
8.70632 
8.71206 

423.565 
425.259 
426.958 

428.661 
430.369 
432.081 

433.798 
4^5.520 
437.245 

1.95830 
1.95917 
1.96004 

1.96091 
1.96177 
1.96264 

1.96350 
1.96437 
1.96523 

4.21904 
4  22091 
4.22278 

4.22465 
4.22651 
4.22838 

4.23024 
4.23210 
4.23396 

9.08904 
9.093(J7 
9.09770 

9.10173 
9.10575 
9.10977 

9.11378 
9.11779 
9.12180 

7.60 

57.7600 

2.75681 

8.71780 

438.976 

1.96610 

4.23582 

9.12581 

7.61 
7.62 
7.63 

7.64 
7.65 
7.66 

7.67 
7.68 
7.69 

57.9121 
58.0644 
58.2169 

58.3096 
58.5225 
58.6756 

58.8289 
58.9824 
59.1361 

2.758(^2 
2.76043 
2.76225 

2.76405 
2.76586 
2.76767 

2.76948 
2.77128 
2.77308 

8.72353 
8.72926 
8.73499 

8.74071 
8.74643 
8.75214 

8.75785 
8.76356 
8.76926 

440.711 
442.451 
444.195 

445.944 
447.697 
449.455 

451.218 
452.985 
454.757 

1.96696 
1.96782 
1.96868 

1.96954 
1.97040 
1.97126 

1.97211 
1.97297 
1.97383 

4.23768 
4.23954 
4.24139 

4.24324 
4.24509 
4.24694 

4.24879 
4.25063 
4.25248 

9.12981 
9.13380 
9.13780 

9.14179 
9.14577 
9.14976 

9.15374 
9.15771 
9.16169 

7.70 

59.2900 

2.77489 

8.77496 

456.533 

1.97468 

4.25132 

9.16566 

7.71 

7.72 
7.73 

7.74 
7.75 
7.76 

7.77 
7.78 
7.79 

59.4441 
59.5984 
69.7529 

59.9076 
60.0625 
60.2176 

60.3729 
60.5284 
60.6841 

2.77C69 
2.77849 
2.78029 

2.78209 
2.78388 
2.78568 

2.78747 
2.78927 
2.79106 

8.78066 
8.78635 
8.79204 

8.79773 
8.80341 
8.80909 

8.81476 
8.82043 
8.82010 

458.314 
460.100 
461.890 

463.685 
465.484 
467.289 

469.097 
470.911 
472.729 

1.97554 
1.97039 
1.97724 

1.97809 
1.97895 
1.97980 

1.98065 
1.98150 
1.98234 

4.25616 
4.25800 
4.25984 

4.20167 
4.26351 
4.26534 

4.26717 
4.26900 

4.27083 

9.16962 
9.17359 
9.17754 

9.18150 
9.18545 
9.18940 

9.19335 
9.19729 
9.20123 

7.80 

60.8400 

2.79285 

8.83176 

474.552 

1.98319 

4.27266 

9.20516 

7.81 
7.82 
7.83 

7.84 

7.85 
7.86 

7.87 
7.88 
7.89 

60.9961 
61.1524 
61.3089 

61.4656 
61.6225 
61.7796 

61.9369 
62.0944 
62.2521 

2.79464 
2.79643 
2.79821 

2.80000 
2.80179 
2.80357 

2.80535 
2.80713 
2.80891 

8.83742 
8.84308 
8.84873 

8.85438 
8.86002 
8.86566 

8.87130 
8.87694 

8.88257 

476.380 
478.212 
480.049 

481.890 
483.737 
485.588 

487.443 
489.304 
491.169 

1.98404 
1.98489 
1.98.573 

1.98658 
1.98742 
1.98826 

1.98911 
1.98995 
1.99079 

4.27448 
4.27031 
4.27813 

4.27995 
4.28177 
4.28359 

4.28540 

4.28722 
4.28903 

9.20910 
9.21302 
9.21695 

9.22087 
9.22479 
9.22871 

9.23262 
9.23653 
9.24043 

7.90 

62.4100 

2.81069 

8.88810 

493.039 

1.99163 

4.29084 

9.24434 

7.91 
7.92 
7.93 

7.94 
7.95 
7.96 

7.97 
7.98 
7.99 

62.5681 
62.7264 
62.8849 

63.0436 
63.2025 
63.3616 

63.5209 
63.6804 
63.8401 

2.81247 
2.81425 
2.81603 

2.81780 
2.81957 
2.82135 

2.82312 
2.82489 
2.82666 

8.89382 
8.89944 
8.90505 

8.91067 
8.91628 
8.92188 

8.92749 
8.93308 
8.93868 

494.914 
496.793 
498.677 

500.566 
502.460 
504.358 

506.262 
508.170 
610.082 

1.99247 
1.99331 
1.99415 

1.99499 
1.9^)582 
1.99666 

1.99750 
1.99833 
1.99917 

4.29265 
4.29446 
4.29627 

4.29807 
4.29987 
4.30168 

4.30348 
4.30528 
4.30707 

9.24823 
9.25213 
9.25602 

9.25991 
9.26380 
9.26768 

9.27156 
9.27544 
9.27931 

n] 


Powers  and  Roots 


XXlll 


n 

n2 

V^ 

VlOn 

n^ 

</h 

<^n 

^100  w 

8.00 

64.0000 

2.82»43 

8.94427 

512.000 

2.00000 

4.30887 

9.28318 

8.01 
8.02 
8.03 

SM 
8.05 
8.06 

8.07 
8.08 
8.09 

64.1601 
64..3204 
64.4809 

64.6416 
64.8025 
64.<)636 

65.1249 
65.2864 
65.4481 

2.83019 
2.831% 
2.83373 

2.8^549 

2.83725 
2.83901 

2.84077 
2.84253 
2.84429 

8.1Hi)86 
8.5)5545 
8.96103 

8.96660 
8.97218 
8.97775 

8.98.332 
8.98888 
8.99444 

513.922 
515.850 
517.782 

519.718 
521.(^60 
523.607 

625.558 
527.514 
529.475 

2.00083 
2.00167 
2.00250 

2.003;« 
2.00416 
2.00499 

2.00582 
2.00664 
2.00747 

4.31066 
4.31246 
4.31425 

4.31604 
4.31783 
4.31961 

4.32140 
4.32318 
4.32497 

9.28704 
9.29091 
9.29477 

9.29862 
9..30248 
9.30633 

9.31018 
9.31402 
9.31786 

8.10 

65.6100 

2.84605 

9.00000 

531.441 

2.00830 

4.32675 

9.32170 

8.11 
8.12 
8.13 

8.14 
8.15 
8.16 

8.17 
8.18 
8.19 

65.7721 
65.9344 
66.0969 

66.2596 
66.4225 
66.5856 

66.7489 
6(5.9124 
67.0761 

2.84781 
2.84956 
2.85132 

2.85307 
2.85482 
2.85657 

2.85832 
2.86007 
2.86182 

9.00555 
9.01110 
9.01665 

9.02219 
9.02774 
9.03327 

9.0.3881 
9.044.34 
9.04986 

5.33.412 
535.387 
537.368 

539.353 
541.343 
543.338 

545.339 
547.343 
549.353 

2.00912 
2.00995 
2.01078 

2.01160 
2.01242 
2.01325 

2.01407 
2.01489 
2.01571 

4.32853 
4.33031 
4.33208 

4.33386 
4.335(53 
4.33741 

4.33918 
4.34095 
4.34271 

9.32553 
9..32936 
9.33319 

9.33702 
9.34084 
9.34466 

9..34847 
9.35229 
9.35610 

8.20 

67.2400 

2.86356 

9.05539 

551.368 

2.01653 

4.34448 

9.35990 

8.21 
8.22 
8.23 

8.24 
8.25 
8.26 

8.27 
8.28 
8.29 

67.4041 
67.5<)84 
67.7329 

67.8976 
68.0625 
68.2276 

68.3929 
68.5584 
68.7241 

2.86531 
2.8<;705 
2.86880 

2.87054 
2.87228 
2.87402 

2.87576 
2.87750 
2.87924 

9.06091 
9.0()642 
9.07193 

9.07744 
9.08295 
9.08845 

9.09395 
9.05)945 
9.10494 

553.388 
555.412 
557.442 

559.476 
561.016 
563.560 

565.609 
569.723 

2.01735 
2.01817 
2.01899 

2.01980 
2.02062 
2.02144 

2.02225 
2.02307 
2.02388 

4.34625 
4.34801 
4.34977 

4.35153 
4.35329 
4.35505 

4.35681 
4.35856 
4.36032 

9.36370 
9..S6751 
9.37130 

9.37510 
9.37889 
9.38268 

9.3864<) 
9.39024 
9.39402 

8.30 

8.31 
8.32 
8.33 

8.34 
8.35 
8.36 

8.37 
8.38 
8.39 

68.8900 

2.88097 

9.11043 

571.787 

2.02469 

4.36207 

9.39780 

69.0561 
69.2224 
69.3889 

69.5556 
69.7225 
69.8896 

70.0569 
70.2244 
70.3921 

2.88271 
2.88444 
2.88617 

2.88791 
2.88964 
2.89137 

2.89310 
2.89482 
2.89655 

9.11592 
9.12140 
9.12688 

9.13236 
9.13783 
9.14330 

9.14877 
9.15423 
9.15%9 

573.856 
575.930 
578.010 

580.094 
582.183 
584.277 

586.376 
588.480 
55X).590 

2.02551 
2.02632 
2.02713 

2.02794 
2.02875 
2.02956 

2.03037 
2.03118 
2.03199 

4.36382 
4.36557 
4.36732 

4..36907 
4.37081 
4.37256 

4.37430 
4.37604 
4.37778 

9.40157 
9.40534 
9.40911 

9.41287 
9.41663 
9.42039 

9.42414 
9.42789 
9.43164 

8  40 

70.5600 

2.89828 

9.16515 

592.704 

2.03279 

4.37952 

9.43539 

8.41 
8.42 
8.43 

8.44 
8.45 
8.46 

8.47 
8.48 
8.49 

70.7281 
70.89«)4 
71.0649 

71.2336 
71.4025 
71. .5716 

71.7409 
71.9104 
72.0801 

2.90000 
2.90172 
2.90345 

2.90517 
2.90689 
2.90861 

2.91033 
2.91204 
2.91376 

9.17061 
9.17606 
9.18150 

9.18695 
9.19239 
9.19783 

9.20326 
9.20869 
9.21412 

594.823 
596.1^8 
599.077 

601.212 
603.ail 
605.496 

607.645 
609.800 
611.960 

2.0.3360 
2.0.3440 
2.03521 

2.03601 
2.03682 
2.03762 

2.03842 
2.03923 
2.04003 

4.38126 
4.38299 
4.38473 

4.38646 
4.38819 
4.38992 

4.39165 
4.39338 
4.39510 

9.43913 
9.44287 
9.44661 

9.45034 
9.454D7 
9.45780 

9.46152 
9.46525 
9.46897 

XXIV 

Powers  and  Roots 

[11 

n 

n^ 

y/n 

n^ 

</n 

^10  n 

VlOfi 

</100n 

8.50 

72.2500 

2.91548 

9.21954 

614.125 

2.04083 

4.39683 

9.47268 

8.51 
8.52 
8.53 

8.54 
8.55 
8.56 

8.57 
8.58 
8.59 

72.4201 
72.5904 
72.7609 

72.9316 
73.1025 
73.2736 

73.4449 
73.6164 

73.7881 

2.91719 
2.91890 
2.92062 

2.92233 
2.92404 
2.92575 

2.92746 
2.92916 
2.93087 

9.22497 
9.23038 
9.23580 

9.24121 
9.24662 
9.25203 

9.25743 
9.26283 
9.26823 

616.295 
618.470 
620.650 

622.836 
625.026 
627.222 

629.423 
631.629 
633.840 

2.04103 
2.04243 
2.04323 

2.04402 
2.04482 
2.04562 

2.04641 
2.04721 
2.04801 

4.39855 
4.40028 
4.40200 

4.40372 
4.40543 
4.40715 

4.40887 
4.41058 
4.41229 

9.47640 
9.48011 
9.48381 

9.48752 
9.49122 
9.49492 

9.49861 
9.50231 
9.50600 

8.60 

73.9600 

2.93258 

9.27362 

636.056 

2.04880 

4.41400 

9.50969 

8.61 
8.62 
8.63 

8.64 
8.65 
8.66 

8.67 
8.68 
8.69 

74.1321 
74.3044 
74.4769 

74.6496 
74.8225 
74.9956 

75.1689 
75.3424 
75.5161 

2.93428 
2.93598 
2.93769 

2.93939 
2.94109 
2.94279 

2.94449 
2.94618 
2.94788 

9.27901 
9.28440 
9.28978 

9.29516 
9.30054 
9.30591 

9.31128 
9.31665 
9.32202 

638.277 
640.504 
642.736 

644.973 

647.215 
649.462 

651.714 
653.972 
656.235 

2.04959 
2.05039 
2.05118 

2.05197 
2.05276 
2.05355 

2.05434 
2.05513 
2.05592 

4.41571 
4.41742 
4.41913 

4.42084 
4.42254 
4.42425 

4.42595 
4.42765 
4.42935 

9.51337 
9.51705 
9.52073 

9.52441 

9.52808 
9.53175 

9.53542 
9.53908 
9.54274 

8.70 

75.6900 

2.94958 

9.32738 

658.503 

2.05671 

4.43105 

9.54640 

8.71 

8.72 
8.73 

8.74 
8.75 
8.76 

8.77 
8.78 
8.79 

75.8641 
76.0384 
76.2129 

76.3876 
76.5625 
76.7376 

76.9129 
77.0884 
77.2641 

2.95127 
2.95296 
2.95466 

2.95635 
2.95804 
2.95973 

2.96142 
2.96311 
2.96479 

9.33274 
9.33809 
9.34345 

9.34880 
9.35414 
9.35949 

9.36483 
9.37017 
9.37550 

660.776 
663.055 
665.339 

667.628 
669.922 
672.221 

674.526 
676.836 
679.151 

2.05750 
2.05828 
2.05907 

2.05986 
2.06064 
2.06143 

2.06221 
2.06299 
2.06378 

4.43274 
4.43444 
4.43613 

4.43783 
4.43952 
4.44121 

4.44290 
4.44459 
4.44627 

9.55006 
9.55371 
9.55736 

9.56101 
9.56466 
9.56830 

9.57194 
9.57557 
9.57921 

8.80 

77.4400 

2.96648 

9.38083 

681.472 

2.06456 

4.44796 

9.58284 

8.81 
8.82 
8.83 

8.84 
8.85 
8.86 

8.87 
8.88 
8.89 

77.6161 
77.7924 
77.9689 

78.1456 
78.3225 
78.4996 

78.6769 
78.8544 
79.0321 

2.96816 
2.96985 
2.97153 

2.97321 
2.97489 
2.97658 

2.97825 
2.97993 
2.98161 

9.38616 
9.39149 
9.39681 

9.40213 
9.40744 
9.41276 

9.41807 
9.42338 
9.42868 

683.798 
686.129 
688.465 

690.807 
693.154 
695.506 

697.864 
700.227 
702.595 

2.06534 
2.06612 
2.06690 

2.06768 
2.06846 
2.06924 

2.07002 
2.07080 
2.07157 

4.44964 
4.45133 
4.45301 

4.45469 
4.45637 
4.45805 

4.45972 
4.46140 
4.46307 

9.58647 
9.59009 
9.59372 

9.59734 
9.60095 
9.60457 

9.60818 
9.61179 
9.61540 

8.90 

79.2100 

2.98329 

9.43398 

704.969 

2.07235 

4.46475 

9.61900 

8.91 
8.92 
8.93 

8.94 
8.95 
8.96 

8.97 
8.98 
8.99 

79.3881 
79.5664 
79.7449 

79.9236 
80.1025 
80.2816 

80.4609 
80.6404 
80.8201 

2.98496 
2.98664 
2.98831 

2.98998 
2.99166 
2.99333 

2.99500 
2.99666 
2.998.33 

9.43928 
9.44458 
9.44987 

9.45516 
9.46044 
9.46573 

9.47101 

9.47629 
9.48156 

707.348 
709.732 
712.122 

714.517 
716.917 
719.323 

721.734 
724.151 
726.573 

2.07313 
2.07390 
2.07468 

2.07545 
2.07622 
2.07700 

2.07777 
2.07854 
2.07931 

4.46642 
4.46809 
4.46976 

4.47142 
4.47309 
4.47476 

4.47642 

4.47808 
4.47974 

9.62260 
9.62620 
9.62980 

9.63339 
9.63698 
9.64057 

9.64415 
9.64774 
9.65132 

n] 


Powers  and  Roots 


XXV 


n 

W2 

yfn 

VIOm 

W8 

^ 

</H}n 

^100  w 

9.00 

81.0000 

3.00000 

0.48083 

729.000 

2.08008 

4.48140 

9.65489 

9.01 
9.02 
9.03 

9.04 
9.05 
9.06 

9.07 
9.08 
9.09 

81.1801 
81.3604 
81.5409 

81.7216 
81.<K)25 
82.083(3 

82.2049 
82.4404 
82.0281 

3.00107 
3.00333 
3.00500 

3.000(56 
3.008.32 
3.00008 

3.01104 
3.01.3:50 
3.01406 

9.49210 
9.49737 
9.50263 

9.50789 
9.5i;il5 
9.51840 

9.523(55 
9.52800 
9.5^415 

731.4.33 
733.871 
736.314 

738.763 
741.218 
743.677 

746.143 
748.613 
751.080 

2.08085 
2.08162 
2.08230 

2.08316 
2.08.303 
2.08470 

2.08546 
2.08023 
2.08(500 

4.48306 
4.48472 
4.48638 

4.48803 
4.480(30 
4.40134 

4.40200 
4.40464 
4.40620 

0.05847 
0.0(5204 
0.06501 

0.66018 
9.07274 
9.07630 

9.67986 
9.68342 
9.68697 

9.10 

82.8100 

3.01002 

9.53930 

753.571 

2.08770 

4.4075)4 

9.69052 

9.11 
9.12 
9.13 

9.14 
9.15 
9.16 

9.17 
9.18 
9.19 

82.9021 
83.1744 
83.3569 

83.5396 
83.7225 
83.9056 

84.0889 
84.2724 
84.4501 

3.01828 
3.01003 
3.02150 

3.02324 
3.024W 
3.02655 

3.02820 
3.02085 
3.03150 

9.54463 
9.54087 
9.55510 

9.50033 
9.50556 
9.57079 

9.57001 
9.58123 
9.58645 

756.058 
758.551 
761.048 

7(53.552 
706.001 
708.575 

771.005 
77:5.(521 
77(5.152 

2.08852 
2.08020 
2.00005 

2.00081 
2.00158 
2.00234 

2.00310 
2.00386 
2.00462 

4.40050 
4.50123 
4.50288 

4.50452 
4.50616 
4.50781 

4.50045 
4.51108 
4.51272 

9.09407 
9.00702 
9.70116 

0.70470 
9.70824 
9.71177 

9.71531 
9.71884 
9.722.36 

9.20 

84.0400 

3.03315 

9.50106 

778.088 

2.00538 

4.51436 

9.72589 

9.21 
9.22 
9.23 

9.24 
9.25 
9.26 

9.27 
9.28 
9.29 

8i.8241 
85.0084 
85.1929 

85.3776 
85.5625 
85.7476 

85.9329 
841.1184 
8(5.3141 

3.03480 
3.03(545 
3.03800 

3.03074 
3.04138 
3.04302 

3.04467 
3.04631 
3.04705 

9.5<)087 
9.00208 
9.00729 

9.01249 
9.01709 
9.02289 

9.62808 
9.03328 
9.(5:3846 

781.2:^,0 
783.777 
780.330 

788.880 
701.4.53 
704.023 

706.508 
700.170 
801.765 

2.00614 
2.0{)(500 
2.00765 

2.00841 
2.00017 
2.00002 

2.10068 
2.10144 
2.10210 

4.51500 
4.517(53 
4.51026 

4.52080 
4.52252 
4.52415 

4.52578 
4.52740 
4.52003 

9.72041 
0.73203 
0.73045 

0.73<iO6 
0.74348 
0.74600 

9.75049 
9.75400 
9.75750 

9.30 

86.4<M)0 

3.04050 

9.64.365 

m\Mi 

2.102<>4 

4.530(55 

9.76100 

9.31 
9.32 
9.33 

9.3i 
9.35 
9.36 

9.37 
9.38 
9.39 

9.40 

9.41 
9.42 
9.43 

9.44 
9.45 
9.46 

9.47 
9.48 
9.49 

8(5.6761 
8(3.8(524 
87.0i89 

87.2356 
87.4225 
87.6006 

87.7969 
87.9844 
88.1721 

3.05123 
3.05287 
3.0.5450 

3.05014 
3.05778 
3.05041 

3.00105 
3.06208 
3.0(5431 

9.64883 
9.65401 
9.65019 

9.60437 
9.(3(3i)54 
9.07471 

9.07988 
9.(58504 
9.(SiK)20 

800.054 
800.5.-)8 
812.166 

814.781 
817.400 
820.026 

822.0,57 
825.2M 
827.0.36 

2.10370 
2.10445 
2.10520 

2.10505 
2.10671 
2.10746 

2.10821 
2.10806 
2.10071 

4.53228 
4.5:3300 
4.53552 

4.53714 
4.53876 
4.54038 

4..54109 
4.54301 
4.54522 

9.7(3450 
9.70799 
9.77148 

9.77497 
9.77846 
9.78105 

9.78543 
9.78891 
9.79239 

88.3600 

3.0(55i)4 

9.605.36 

830.584 

2.11045 

4.54684 

9.79586 

88.5481 
88.7:364 
88.9249 

89.113(5 
89.3025 
89.4916 

89.6809 
80.8704 
90.0001 

3.0(5757 
3.00020 
3.07083 

3.07246 
3.07409 
3.07571 

3.07734 
3.07806 
3.08058 

9.70052 
9.70567 
9.71082 

9.71597 
9.72111 
9.72025 

9.73139 
9.736.53 
9.74166 

8.33.238 
835.807 
838.562 

841.232 
843.000 
846.501 

840.278 
851.W1 
854.670 

2.11120 
2.11105 
2.11270 

2.11344 
2.11410 
2.114^ 

2.11568 
2.11642 
2.11717 

4.54845 
4.55006 
4.55167 

4.55328 
4.55488 
4.55649 

4.55809 
4.55{>70 
4.56130 

9.79933 
9.80280 
9.80627 

9.80974 
9.81320 
9.81666 

9.82012 
9.82:357 
9.82703 

xxvi 

Powers  a 

nd  Roots 

[11 

n 

^2 

Vn 

VlOn 

n^ 

^n 

</10n 

S/lOOn 

9.50 

90.2500 

3.08221 

9.74679 

857.375 

2.11791 

4.r,6290 

9.83048 

9.51 
9.52 
9.53 

9.54 
9.55 
9.56 

9.57 
9.58 
9.59 

90.4401 
90.6304 
90.8209 

91.0116 
91.2025 
91.3936 

91.5849 
91.7764 
91.9()81 

3.08383 
3.08545 
3.08707 

3.08869 
3.09031 
3.09192 

3.09354 
3.09516 
3.09677 

9.75192 
9.75705 
9.76217 

9.76729 
9.77241 
9.77753 

9.78264 
9.78775 
9.79285 

860.085 
862.801 
865.523 

868.251 
870.984 
873.723 

876.467 
879.218 
881.974 

2.118()5 
2.11940 
2.12014 

2.12088 
2.12162 
2.12236 

2.12310 
2.12384 
2.12458 

4.56450 
4.56(510 
4.56770 

4.56930 
4.57089 
4.57249 

4.57408 
4.57567 

4.57727 

9.83392 
9.83737 
9.84081 

9.84425 
9.84769 
9.85113 

9.85456 
9.85799 
9.86142 

9.60 

92.1600 

3.09839 

9.79796 

884.736 

2.12532 

4.57886 

9.86485 

9.61 
9.62 
9.63 

9.64 
9.65 
9.66 

9.67 
9.68 
9.69 

92.3521 
92.5444 
92.7369 

92.9296 
93.1225 
93.3156 

93.5089 
93.7024 
93.8961 

3.10000 
3.10161 
3.10322 

3.10483 
3.10644 
3.10805 

3.10966 
3.11127 
3.11288 

9.80306 
9.80816 
9.81326 

9.81835 
9.82344 
9.82853 

9.83362 
9.83870 
9.84378 

887.504 
890.277 
893.056 

895.841 
898.632 
901.429 

904.231 
907.039 
909.853 

2.12605 
2.12679 
2.12753 

2.12826 
2.12900 
2.12974 

2.13047 
2.13120 
2.13194 

4.58045 
4.58204 
4.58362 

4.58521 

4.C8()79 
4.58838 

4.58996 
4.59154 
4.59312 

9.86827 
9.87169 
9.87511 

9.87853 
9.88195 
9.88536 

9.88877 
9.89217 
9.89558 

9.70 

94.0900 

3.11448 

9.84886 

912.673 

2.13267 

4.59470 

9.89898 

9.71 
9.72 
9.73 

9.74 
9.75 
9.76 

9.77 
9.78 
9.79 

94.2841 
94.4784 
94.6729 

94.8676 
95.0625 
95.2576 

95.4529 
95.6484 
95.8441 

3.11609 
3.11769 
3.11929 

3.12090 
3.122.50 
3.12410 

3.12570 
3.12730 
3.12890 

9.85393 
9.85901 
9.86408 

9.86914 

9.87421 
9.87927 

9.88433 
9.88939 
9.89444 

915.499 
918.330 
921.167 

924.010 

926.859 
929.714 

932.575 
935.441 
938.314 

2.13340 
2.13414 
2.13487 

2.13560 
2.13633 
2.13706 

2.13779 

2.138.52 
2.13925 

4.59(528 
4.59786 
4.59943 

4.60101 
4.60258 
4.60416 

4.60573 
4.60730 

4.60887 

9.90238 
9.90578 
9.90918 

9.91257 
9.91596 
9.91935 

9  92274 
9.92612 
9.92950 

9.80 

96.0400 

3.13050 

9.89949 

941.192 

2.13997 

4.61044 

9.93288 

9.81 
9.82 
9.83 

9.84 
9.85 
9.86 

9.87 
9.88 
9.89 

96.2361 
96.4324 
96.6289 

96.8256 
97.0225 
97.2196 

97.4169 
97.6144 
97.8121 

3.13209 
3.13369 
3.13528 

3.13688 
3.13847 
3.14006 

3.14166 
3.14325 
3.14484 

9.90454 
9.90959 
9.91464 

9.91968 
9.92472 
9.92975 

9.93479 
9.93982 
9.94485 

944.076 
946.966 
949.862 

952.764 
955.672 
958.585 

961.505 
964.430 
967.362 

2.14070 
2.14143 
2.14216 

2.14288 
2.14361 
2.14433 

2.14506 
2.14578 
2.14651 

4.61200 
4.61357 
4.61514 

4.61670 
4.61826 
4.61983 

4.62139 
4.62295 
4.62451 

9.93626 
9.93964 
9.94301 

9.f)4638 
9.94975 
9.96311 

9.95648 
9.95984 
9.96320 

9.90 

98.0100 

3.14643 

9.94987 

970.2i)9 

2.14723 

4.62607 

9.9(5655 

9.91 
9.92 
9.93 

9.94 
9.95 
9.96 

9.97 
9.98 
9.9<) 

98.2081 
98.4064 
98.6049 

98.8036 
99.0025 
99.2016 

99.4009 
99.6004 
99.8001 

3.14802 
3.14<)60 
3.15119 

3.15278 
3.15436 
3.15595 

3.15753 
3.15911 
3.16070 

9.95490 
9.95992 
9.96494 

9.96995 
9.97497 
9.97998 

9.98499 
9.98999 
9.99500 

973.242 
976.191 
979.147 

982.108 
985.075 
988.048 

9^)1.027 
f)94.012 
997.003 

2.14795 

2.14807 
2.14940 

2.15012 
2.1.5084 
2.15156 

2.15228 
2.15300 
2.15372 

4.62762 
4.62918 
4.63073 

4.63229 
4.63384 
4.63539 

4.63694 
4.63849 
4.64004 

9.96991 
9.9732(5 
9.97661 

9.97996 
9.98331 
9.98665 

9.989i)9 
9.99333 
9.99(567 

TABLE   III  —  IMPORTANT   NUMBERS 


A,     Units  of  Length 


English  Units 


Metric  Units 


12  inches  (in.)  =  1  foot  (ft.)  10  millimeters  =  1  centimeter  (cm.) 

3  feet  =  1  yard  (yd.)  (mm.) 

6^  yards  =  1  rod  (rd.)  10  centimeters  =  1  decimeter  (dm.) 

320 rods  =1  mile  (mi.)        10  decimeters    =  1  meter  (m.) 

10  meters  =  1  dekameter  (Dm.) 

1000  meters      =  1  kilometer  (Km.) 

Metric  to  English 

1  cm.    =  0.3937  in. 

Im.     =39.87  in.  =3.2808  ft. 
1  Km.  =  0.6214  mi. 


English  to  Metric 

1  in.   =  2.5400  cm. 

1  ft.    =  30.480  cm. 
1  mi.  =  1.6093  Km. 


1  square  mile  =  640  acres 


Units  of  Area  or  Surface 

feet  =  1296  square  inches 
-'-'-  =  4840  square  yards 
=  102400  square  rods 


B. 

1  square  yard  =     9  square  icci,  —  i^uyj  04110,1^  mv^nw 
1  acre  (A.)      =  160  square  rods  =  4840  square  yards 


C.     Units  of  Measurement  of  Capacity 

Liquid  Measure 


Dry  Measure 
2  pints  (pt.)  =  1  quart  (qt.) 
8  quarts  =  1  peck  (pk.) 
4  pecks  =  1  bushel  (bu.) 


4  gills  (gi.)  =  1  pint  (pt.) 
2  pints  =  1  quart  (qt.) 

4  quarts       =  1  gallon  (gal.) 
1  gallon       =  231  cu.  in. 


D.    Metric  Units  to  English  Units 

1  liter  =  1000  cu.  cm.  =  61.02  cu.  in.  =  1.0567  liquid  quarts 
1  quart  =  .94636  liter  =  946.36  cu.  cm. 
1000  grams  =  1  kilogram  (Kg.)  =  2.2046  pounds  (lb.) 
1  pound  =  .463593  kilogram  =  453.59  grams 


E,     Other  Numbers 

T  =  ratio  of  circumference  to  diameter  of  a  circle 
=  3.14159265 
1  radian  =  angle  subtended  by  an  arc  equal  to  the  radius 

=  57^  17'  44".8  =  57°.2957795  =  18077r 
1  degree  =  0.01745329  radian,  or  7r/180  radians 
Weight  of  1  cu.  ft.  of  water  =  62.425  lb. 
xxvii 


INDEX 

[Numbers  refer  always  to  pages.] 


Acute  angle,  15. 

Acute  triangle,  35. 

Adjacent  angles,  17. 

Alternate  exterior  angle,  49. 

Alternate  interior  angle,  49. 

Alternation,  proportion  by,  128. 

Altitude,  of  a  parallelogram,  69 ;  of 
a  trapezoid,  69 ;    of  a  triangle,  36. 

Angle,  6,  11,  15;  acute,  15;  central, 
89,  112;  of  elevation,  20;  exterior, 
21,  76;  generation  of,  14;  in- 
cluded, 21;  inscribed,  112;  in- 
terior, 21,  76;  measure  of,  112; 
obtuse,  15;  right,  15;  straight,  15; 
units  of,  12,  15,  202 ;  vertex  of,  11. 

Angles,  adjacent,  17 ;  alternate-ex- 
terior, 49 ;  alternate-interior,  49  ; 
complementary,  17 ;  correspond- 
ing, 49;  exterior,  49;  interior,  49; 
vertical,  17. 

Angular  speed,  97. 

Antecedents  of  a  proportion,  126. 

Apothem,  189. 

Arc,  1,  89;  intercepted,  89;  major, 
93;  minor,  93. 

Area,  22,  162;  of  a  circle,  191,  194; 
of  a  circular  ring,  196 ;  of  a  field, 
181 ;  of  a  parallelogram,  164 ;  of  a 
rectangle,  22,  162 ;  of  a  regular 
polygon,  185,  189  ;  of  a  sector,  195 ; 
of  a  trapezoid,  168;  of  a  triangle, 
166,  180,  184,  185. 

Aristotle,  58. 

Assumptions,  26. 

Axioms,  lists  of,  26. 

Axiom,  26;  of  measurement,  107. 
See  also  Postulate. 

3ase,   of  a  quadrilateral,   68;    of  a 

trapezoid,  68 ;  of  a  triangle,  36. 
Bhaskara,  183. 


Center,  of  a  circle,  1,  21,  89;  of  p. 
regular  polygon,  189. 

Central  angle,  89,  110,  112. 

Chord,  89,  148. 

Circle,  1,  21,  89  ;  area  of,  191,  194  ;  arc 
of,  1,  89;  center  of,  1,  21,  89;  cir- 
cumscribed, 99,  191 ;  circumference 
of,  1,  21,  89,  191,  193  ;  diameter  of, 
89;  inscribed,  99,  191 ;  length  of, 
191,  192;  radius  of,  1,  21,  89. 

Circles,  concentric,  89. 

Circular  ring,  196. 

Circumcenter,  105. 

Circumference  of  a  circle,  1,  21,  89, 
191,  192. 

Circumscribed  figures,  99,  187,  191. 

Commensurable  quantities,  108. 

Compasses,  1,  19. 

Complement,  17. 

Complementary  angles,  17. 

Complete  revolution,  15. 

Composition,  proportion  by,  128. 

Composition  and  division,  propor- 
tion by,  128. 

Concentric  circles,  89. 

Conclusion,  44. 

Congruent  figures,  36. 

Congruence  of  triangles,  37. 

Consequents  of  a  proportion,  126. 

Construction  of  figures,  19. 

Construction  problems,  preliminary, 
28. 

Contact,  point  of,  99. 

Continued  proportion,  126. 

Converse  of  a  theorem,  52. 

Convex  polygon,  76. 

Corollary,  27. 

Corresponding  angles,  49. 

Cosine  of  an  angle,  153. 

Cosine  law,  177. 

Curve,  10 ;  generation  of  a,  14. 


xxlx 


XXX 


INDEX 


Decagon,  76. 

Degree,  12,  91. 

Diagonal,  of  a  rectangle,  22;  of  a 
quadrilateral,  69  ;  of  a  polygon,  76. 

Diameter,  89. 

Diamond,  69. 

Division,  in  a  proportion,  128;  con- 
struction for,  158. 

Drawing  triangle,  41,  57. 

Equiangular,   polygon,   76;   triangle, 

35;   triangles,  139. 
Equilateral,  polygon,  76 ;  triangle,  35. 
Equivalent,  162. 
Enlargement,  19,  139,  142,  159. 
Euclid,  49,  183. 
Exterior  angle,  21,  49,  76. 
Extremes  of  a  proportion,  126. 

Fourth  proportional,  137. 

Hexagon,  76. 
Hypotenuse,  36. 
Hypothesis,  44. 

Included  angle,  21. 

Incommensurable  quantities,  108. 

Indirect  proof,  51. 

Inscribed  angle,  112. 

Inscribed  figures,  99,  191. 

Intercept,  89. 

Interior  angle,  21,  49,  76. 

Inversion  of  a  proportion,  127. 

Inversely  proportional,  157. 

Isoperimetric  figures,  208 

Isosceles,  trapezoid,  87;  triangle,  35. 

Kite,  87. 

liength,  of  curves,  23  ;  of  a  circle,  191, 

193. 
Limit,  109. 
Locus,  79. 

Maximum,  206. 

Mean  proportional,  144. 

Means  of  a  proportion,  126. 

Measure,   of   central   angle,    112;   of 

inscribed    angle,    1 12 ;    numerical, 

107. 


Measurement,  107. 
Median,  of  trapezoid,  168 ;  of  a  tri- 
angle, 36. 
Minimum,  206. 
Minute,  12. 
Multiplication,  construction  for,  159. 

Obtuse  angle,  15. 
Obtuse  triangle,  35. 
Octagon,  76. 

Parallel  lines,  49. 

Parallelogram,  57,  68 ;   area  of,  164. 

Pentagon,  76. 

Perigon,  15. 

Perimeter,  189. 

Perpendicular,  4,  15. 

Pi  (tt),  193,  198. 

Point,  10 ;  of  contact,  99 ;  of  tan- 
gency,  99. 

Polygon,  76 ;  circumscribed,  99,  191 ; 
convex,  76  ;  equiangular,  76  ;  equi- 
lateral, 76  ;  inscribed,  99,  191 ;  reg- 
ular, 76,  185,  186,  189;  similar, 
139,  146. 

Postulate,  26;  on  areas,  191;  on 
central  angles,  90 ;  on  geometric 
ratios.  111 ;  on  lengths,  191 ;  on 
parallels,  49  ;   Hst  of,  27. 

Problem,  27. 

Projection,  175. 

Proportion,  126. 

Proportional.  See  Fourth,  Mean, 
Third. 

Proposition,  27. 

Protractor,  12. 

Pythagoras,  58,  173,  183. 

Pythagorean  theorem,  172. 

Quadrant,  89. 
Quadrilateral,  22,  68,  76. 

Radian,  202. 

Radius,  of  a  circle,  1,  21,  89;    of   a 

regular  polygon,  189. 
Ratios,  107,  126;   postulate  of.  111; 

trigonometric,  153. 
Rectangle,  22,  68 ;  area  of,  22,  162. 
Rectilinear  figures,  36. 
Reductio  ad  absurdum,  51. 


INDEX 


XXXI 


Reduction,  19,  139.  See  also  Enlarge- 
ment. 

Reduction  to  an  absurdity,  51. 

Regular  polygon,  76,  186;  apothem 
of,  189 ;  area  of,  185,  189 ;  radius 
of,  189. 

Revolution,  complete,  15.  See  also 
Rotation. 

Rhombus,  68. 

Right  angle,  15. 

Right  triangle,  36. 

Ring,  circular,  196. 

Rotation,  14,  97. 

Ruler,  1,  19. 

Scalene  triangle,  35. 

Secant,  99,  148. 

Second,  12. 

Sector,  89  ;   area  of,  195. 

Segment,  of  a  circle,  89  ;  of  a  line,  36  ; 

proportional,  131. 
Side,  of  an  angle,  11;    of  a  polygon, 

76 ;    of  a  right  triangle,  36 ;  of  a 

triangle,  21. 
Similarity,  139. 
Similar  polygons,  139,  146. 
Similar  triangles,  139. 
Sine  law,  161. 
Sine  of  an  angle,  153. 
Solid,  10 ;  generation  of,  14. 
Speed,  angular,  97. 
Square,  9,  22,  68. 
Squared  paper,  22. 


Square  root,  construction  for,  160. 
Straight  angle,  15. 
Subtend,  89. 
Supplement,  17. 
Supplementary  angles,  17. 
Surface,  10 ;   generation  of,  14. 
Symbols,  list  of,  34. 

Tables,  i-xxvii ;  of  chords,  96  ;  trig- 
onometric, 153,  155. 

Tangent  of  an  angle,  153. 

Tangent  to  a  circle,  99,  101,  148. 

Terms  of  a  proportion,  126. 

r hales,  41. 

Theorem,  27. 

Theorems,  preliminary,  28. 

Third  proportional,  137. 

Transversal,  49. 

Trapezoid,  68;  area  of,  168;  isosceles, 
87. 

Triangle,  2,  21,  35,  36,  76;  area  of, 
166,  180,  184,  185;  acute,  .35; 
altitude  of,  36  ;  circumscribed,  99  ; 
equiangular,  35  ;  equilateral,  35  ; 
inscribed,  99  ;  isosceles,  35  ;  obtuse, 
35  ;   median  of,  36  ;   right,  36. 

Triangles,  equiangular,  139 ;  similar, 
139,  152 ;   solution  of,  152. 

Trigonometry,  152. 

Vertex,  of  an  angle,  11 ;  of  a  triangle, 

21. 
Vertical  angles,  17. 


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